EE 66 Exam 1 — Complete Concept Map (Fa26)

Built from Fa24, Sp25, Fa25, Sp26 Exam 1 papers + Fa26 HW1–HW2 and Discussions 1A, 1B, 2A, 2B. Thirteen concepts. Each section: core theory, how it's actually been tested, traps, and related topics that deepen the concept. This is the spec for the practice sheet — one block of 6 problems per concept.

Exam profile (all four iterations)

70 min design / 80 min hard cap, closed book, one double-sided handwritten cheat sheet allowed.
Every paper ships a Potentially Useful Facts page: inner product definitions, Cauchy-Schwarz, triangle inequality, geometric sum formula, angle formula, $\sum k = \frac{n(n+1)}{2}$, $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$, polynomial root facts, common trig values. You don't memorize these — you drill deploying them fast.
~15 free points (name/SID on every page + hand-copied pledge). Take them.
Grading language never changes: "succinct, yet clear and convincing." Every claim gets one line of justification. "Reasonably simple expression" = closed form — a final answer with a leftover $\sum$ or $\int$ loses points.
~180–225 points over 70 min ≈ 3 pts/min. A 10-point part deserves ~3 minutes.
You may use the result of part (a) in part (b) even if you couldn't prove (a). Never skip a chain because the first link broke.
Trend that matters: Fa24/Sp25 lived in $\mathbb{R}^n$ + polynomials. Fa25/Sp26 shifted hard into signals — DT/CT signals as vectors, integral/sum inner products, complex exponentials. Expect Fa26 to dress the same linear algebra in signal clothing.

Priority tiers for Fa26

Tier 1 — every-year locks: subspace test proofs (C4), independence/basis (C6–C7), norm computations (C9), inner-product expansion identities (C8, C11).
Tier 2 — recent-trend heavy: complex exponentials / roots of unity (C3, C12), signal norms & inner products via sums/integrals (C8–C9), and the concrete signal toolbox (C13) — deltas, steps, parity: the raw material Fa25/Sp26-style problems are built from.
Tier 3 — HW2-flagged, fresh this semester: subspace algebra $S+T$, $S\cap T$, dimension formula (C5); projection formula (C10). Past exams never isolated these; your homework spent two full problems on them. For a course that rewrites itself each term, that's a strong signal.
Tier 4 — occasional: complex-plane loci sketches (C2, twice), nearest-neighbor geometry (C10, once).
Timing tell: Dis 2B — the most recent discussion before your exam window — re-teaches the exact Fa24 MT1.3 machinery (Gram matrix, orthonormality, expansion coefficients, projection) on the same four DFT vectors. Staff walking through an old exam problem right before the exam is as loud as signals get: treat C10's expansion block as near-certain.

C1. Complex numbers: Cartesian ↔ polar, Euler's formula

Core theory

Two coordinate systems for the same point:

Cartesian: $z = a + ib$, with $a = \operatorname{Re}(z)$, $b = \operatorname{Im}(z)$.
Polar: $z = re^{i\theta}$, with $r = |z| = \sqrt{a^2+b^2}$ and $\theta$ the angle from the positive real axis. Conversions: $a = r\cos\theta$, $b = r\sin\theta$.
Euler: $e^{i\theta} = \cos\theta + i\sin\theta$. Anchors: $e^{i0}=1$, $e^{i\pi/2}=i$, $e^{i\pi}=-1$, $e^{i3\pi/2}=-i$, $e^{i\pi/4} = \frac{\sqrt2}{2}(1+i)$.

Which form for which job:

Add/subtract → Cartesian (componentwise).
Multiply/divide/power → polar: moduli multiply, angles add. $z^n = r^n e^{in\theta}$ (De Moivre). This is why $z = e^{i\pi/4}$ raised to powers just walks around the unit circle.

Fluency set (know cold):

$\frac{1}{i} = -i$ (multiply top and bottom by $-i$, or by $i^*$).
$(1+i)^2 = 1 + 2i + i^2 = 2i$. Faster in polar: $(\sqrt2 e^{i\pi/4})^2 = 2e^{i\pi/2} = 2i$.
$i^4 = 1$; generally $i^k$ cycles with period 4.
Division: $\frac{z}{w} = \frac{zw^*}{|w|^2}$.
Inverse Euler (Dis 2B Q2 — derive it, don't just quote it): add/subtract $e^{i\theta} = \cos\theta + i\sin\theta$ and $e^{-i\theta} = \cos\theta - i\sin\theta$: $$\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \qquad \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}.$$ These convert trig ↔ exponentials in both directions — the standard move for turning sinusoids into complex-exponential algebra.
De Moivre, derived in one line: $(\cos\theta + i\sin\theta)^n = (e^{i\theta})^n = e^{in\theta} = \cos n\theta + i\sin n\theta$.

How it's been tested

Fa24 MT1.2(a): express $\frac1i$, $(1+i)^2$, $i^4$ in Cartesian form.
Sp26 E1.1(a): Cartesian form of $z = e^{i\pi/4}$.
Dis 2B Q2: derive inverse Euler and De Moivre from Euler's formula.

Traps

Sign of $\frac1i$ — it is $-i$, not $i$.
Quadrant of $\theta$: $-1-i$ has $\theta = 5\pi/4$ (or $-3\pi/4$), not $\pi/4$.
Answer in the demanded form. "Cartesian" means $a+ib$ with real $a,b$ — leaving $2e^{i\pi/2}$ loses points.

De Moivre for $\cos(n\theta)$, $\sin(n\theta)$ identities.
Phasors: every sinusoid $A\cos(\omega t + \phi)$ is $\operatorname{Re}{Ae^{i\phi}e^{i\omega t}}$ — where this course's signal story is heading.

C2. Conjugates, modulus, and loci in the complex plane

Core theory

Conjugate: $z^* = a - ib = re^{-i\theta}$ (reflect across the real axis).
Identities to wield automatically: $zz^* = |z|^2$; $\operatorname{Re}(z) = \frac{z+z^*}{2}$; $\operatorname{Im}(z) = \frac{z-z^*}{2i}$; $(zw)^* = z^*w^*$; $|zw| = |z||w|$.
Geometric dictionary:
- $|z - c| = r$: circle of radius $r$ centered at $c$. With $\le$: closed disk.
- $|z - a| = |z - b|$: perpendicular bisector of the segment from $a$ to $b$ ("equidistant from two points"). With $\le$: the closed half-plane containing $a$.
- Always factor constants out of the modulus first: $|2z - 4| \le 2 \iff 2|z-2| \le 2 \iff |z - 2| \le 1$ — a disk at $2$, radius $1$, not radius 2 at 4.
Symbolic conditions — two attack modes:
- Cartesian substitution: plug $z = a+ib$, separate real and imaginary parts.
- Polar/conjugate algebra: e.g. $z = |z|$ forces $\theta = 0$ (or $z=0$) → the nonnegative real axis. $z^2 = (z^*)^2 \iff (z-z^*)(z+z^*) = 0 \iff \operatorname{Im}(z)=0$ or $\operatorname{Re}(z)=0$ → the union of both axes. (Alternative: ${} z^2 = (z^2)^* {}$ means $z^2$ is real, so $2\theta$ is a multiple of $\pi$.)

How it's been tested

Fa24 MT1.2(b): sketch $|2z-4| \le 2$ and $|z-i| \le |z+i|$ (lower half-plane incl. real axis).
Sp25 MT1.6(a),(b): describe ${z : z=|z|}$ and ${z : z^2 = (z^*)^2}$.

Traps

Boundary: $\le$ includes the circle/line — say so and shade accordingly.
Shading the wrong side of a bisector. Verify by testing one easy point (test $z=i$ against $|z-i| \le |z+i|$: $0 \le 2$ ✓, so shade the side containing $i$... careful — for that inequality the region is the points closer to $i$... no: $|z-i|\le|z+i|$ means closer to $i$ than to $-i$, i.e. the upper half-plane. Always re-derive; never pattern-match from memory).
Forgetting $z = 0$ edge cases when dividing by $z$ or canceling moduli.

The bisector fact is exactly the decision boundary in nearest-neighbor classification (C10) — same math, real coordinates.
$|z - c| \le r$ regions reappear later as stability/ROC-style disk conditions in systems courses.

C3. Powers of $i$, roots of unity, geometric sums

Core theory

$i^k$ cycles $1, i, -1, -i$ with period 4, and any four consecutive powers sum to zero. For $\sum_{k=0}^{N} i^k$: count the terms ($N+1$ of them — off-by-one is the killer), strip complete cycles of 4 (they contribute 0), sum the remainder.
- $\sum_{k=0}^{1000} i^k$: 1001 terms = 250 cycles + 1 leftover → $i^0 = 1$.
- $\sum_{k=0}^{999} i^k$: 1000 terms = 250 cycles → $0$.
$N$-th roots of unity: the $N$ solutions of $z^N = 1$ are $z_k = e^{i2\pi k/N}$, equally spaced on the unit circle. $z = e^{i\pi/4}$ is a primitive 8th root; $z^n$ walks counterclockwise in steps of $45°$, returning after 8 steps.
Their sum is zero — two proofs, know both:
- Algebraic: geometric sum with ratio $z \ne 1$: $\sum_{n=0}^{N-1} z^n = \frac{z^N - 1}{z - 1} = \frac{1-1}{z-1} = 0$.
- Geometric: the points are symmetric about the origin (for even $N$, they cancel in diametrically opposite pairs; in general the configuration's center of mass is the origin).
Geometric sum formula (given on the facts sheet): $\sum_{\ell=M}^{N} \alpha^\ell = \frac{\alpha^{N+1}-\alpha^M}{\alpha - 1}$ for $\alpha \ne 1$, and $N - M + 1$ when $\alpha = 1$. The $\alpha = 1$ branch is not decoration — it's a graded case.

How it's been tested

Fa24 MT1.2(a)(iv): show $\sum_0^3 i^k = 0$, evaluate $\sum_0^{1000} i^k$.
Sp25 MT1.6(c): $\sum_0^{999} i^k$.
Sp26 E1.1(b),(c): plot all 8 values of $e^{i\pi n/4}$ on the unit circle; explain $\sum_0^7 x[n] = 0$ geometrically or algebraically.

Traps

Term counting: $\sum_{k=0}^{N}$ has $N+1$ terms. Every wrong answer in this family is an off-by-one.
When plotting the 8 roots, label which point is which $n$ — the diagram is graded on labels.
Using the geometric sum formula with $\alpha = 1$ in the $\alpha \ne 1$ branch (division by zero).

The Fa24 MT1.3 vectors $\varphi_k$ are literally the columns of the 4-point DFT matrix — mutually orthogonal complex exponential vectors. Roots-of-unity sums are the engine behind DFT orthogonality and Fourier series, the destination of this course.

C4. Vector spaces and the subspace test

Core theory

A vector is anything you can add and scale consistently: arrows in $\mathbb{R}^n$/$\mathbb{C}^n$, polynomials, CT signals $x:\mathbb{R}\to\mathbb{C}$, DT signals $x:\mathbb{Z}\to\mathbb{C}$. The zero vector is the identically-zero object (the all-zeros function, not the number 0 — say this precisely in proofs).

Subspace test. $S \subseteq V$ is a subspace iff:

$\mathbf{0} \in S$ (fastest kill shot — check first),
closed under scalar multiplication: $v \in S, \alpha \in F \Rightarrow \alpha v \in S$,
closed under addition: $v, w \in S \Rightarrow v + w \in S$.

Proof pattern (yes): take generic elements ("let $x, y \in S$, so they satisfy constraint …"), show the constraint survives scaling and addition, write one line per axiom. Disproof pattern (no): one explicit, concrete counterexample violating one axiom. Never hand-wave a "no."

The structural shortcut: sets carved out by homogeneous linear constraints are subspaces; constraints that are inhomogeneous ($= 1$), nonlinear, or about magnitude/shape usually fail.

Verdict catalog (all from these exams/HW — this table is cheat-sheet material):

Set	Verdict	Why
${x \in \mathbb{R}^n : \frac1n\sum x_k = 0}$ (zero mean)	✅	homogeneous linear constraint
CT signals with $x(t)=0$ for $	t	>1$ (limited support)
DT signals with $x[0] = 1$	❌	zero signal not in the set
Bounded signals $\ell^\infty(\mathbb{Z})$	✅	$
Finite-energy signals $\ell^2(\mathbb{Z})$	✅	addition closure via triangle ineq.: $\\|x+y\\| \le \\|x\\|+\\|y\\| < \infty$
Solutions of $\frac{dy}{dt} + \alpha y = 0$, $\alpha$ fixed	✅	linearity of $\frac{d}{dt}$: $(y_1+y_2)' + \alpha(y_1+y_2) = 0$
Same, but "for some $\alpha$" (union over all $\alpha$)	❌	$e^{-t} + e^{-2t}$ solves no single such ODE
Polynomials in $P_n$ with $v(0)=v(1)=0$	✅	two homogeneous linear constraints
${re^{i\pi/3} : r \in \mathbb{R}} \subset \mathbb{C}\cong\mathbb{R}^2$ (line through origin)	✅	1-D line; basis ${(\tfrac12, \tfrac{\sqrt3}{2})}$
${3e^{i\theta}}$ (circle radius 3)	❌	no zero; not closed under scaling
${a + 2i : a \in \mathbb{R}}$ (shifted line)	❌	no zero
Purely imaginary axis ${bi}$	✅	line through origin
${(2(x+y),,x,,y) : x,y \in \mathbb{R}}$	✅	a span in disguise: $x(2,1,0)^\top + y(2,0,1)^\top$ — and every span is a subspace
${(x,,y,,z+1) : x,y,z \in \mathbb{R}}$	✅	the offset is absorbable: $z+1$ sweeps all of $\mathbb{R}$, so the set is $\mathbb{R}^3$
${(x,,y,,x+1) : x,y \in \mathbb{R}}$	❌	genuine inhomogeneous relation $x_3 = x_1 + 1$; the zero vector fails it

Parametrized sets — judge the set, not the letters (HW1 Q4). A set written as ${(\text{expressions in } x, y, \dots)}$ is the image of a parametrization. First try to rewrite it as a span (span ⇒ subspace, done). Then check whether an apparent offset is absorbable by a free parameter before declaring failure: $(x, y, z+1)$ over all $x, y, z$ is $\mathbb{R}^3$, while $(x, y, x+1)$ is a genuinely shifted plane. The set is what matters, not how it's spelled.

The quantifier is the whole game in the ODE row: "for a fixed $\alpha$" gives one solution space (subspace); "for some $\alpha$" gives a union of solution spaces, and unions of distinct subspaces (neither containing the other) are essentially never subspaces — they fail additive closure. Read the set-builder notation aloud before answering.

How it's been tested

Every single exam. Fa24 MT1.4(b) (polynomial roots), Fa25 E1.4 — 60 points, five subspace verdicts in a row, Sp26 E1.2(b) ($\ell^2$ closure), HW2 Q1–Q2, HW1 Q4 (parametrization traps), Dis 1B Q1(a).

Traps

Skipping the explicit counterexample on a "no."
Scalar field mismatch: if the space is over $\mathbb{C}$, closure must hold for complex scalars.
For $\ell^2$ addition closure, "sum of finite is finite" is not an argument — you need $|x+y| \le |x| + |y|$ or $(a+b)^2 \le 2a^2 + 2b^2$.
Writing "$0 \in S$" without saying which zero (the zero function/signal) and why it satisfies the constraint.

Every homogeneous-constraint set is the null space of a linear map — the unifying reason the shortcut works.
Affine sets (subspace + offset, like ${a+2i}$) — everything a subspace is, minus the origin. Solution sets of _in_homogeneous systems.

C5. Subspace algebra: $S+T$, $S \cap T$, dimension formula

Core theory

For subspaces $S, T \subseteq V$:

Sum: $S + T := {s + t : s \in S, t \in T}$ — a subspace. Closure proof shape: $(s_1+t_1) + (s_2+t_2) = (s_1+s_2) + (t_1+t_2)$, regroup into an $S$-part plus a $T$-part using closure of each.
Intersection: $S \cap T$ — a subspace. Proof shape: $v, w \in S\cap T$ means each is in $S$ and in $T$; apply closure inside each space separately.
Union: $S \cup T$ — generally not a subspace (two distinct lines through the origin: the sum of one vector from each leaves the union). Only works when one contains the other.
Direct-sum case, $S \cap T = {0}$: if $B_S$, $B_T$ are bases, then $B_S \cup B_T$ is a basis for $S+T$:
- Spans: any $s + t$ expands in $B_S$ then $B_T$.
- Independent: if $\sum a_i s_i + \sum b_j t_j = 0$, then $\underbrace{\sum a_i s_i}_{\in S} = \underbrace{-\sum b_j t_j}_{\in T} \in S \cap T = {0}$, so both halves are zero, so all coefficients vanish by independence of each basis.
- Hence $\dim(S+T) = \dim S + \dim T$.
General dimension formula: $\dim(S+T) = \dim S + \dim T - \dim(S \cap T)$ (inclusion–exclusion for dimensions; the overlap would otherwise be counted twice).

How it's been tested

HW2 Q2, head-on, all four parts — new to this semester's emphasis; no past Exam 1 isolated it. Fresh homework material that past exams lack is exactly what a rewritten exam mines.

Traps

In the basis-union independence proof, the step "$s = -t \in S\cap T = {0}$" is the crux — without $S\cap T = {0}$ it fails, and the union of bases can be dependent.
$S + T$ is not $S \cup T$. The sum is usually much bigger.
Sanity check the formula: two distinct planes through the origin in $\mathbb{R}^3$: $2 + 2 - \dim(\text{line}) = 3$. ✓

Direct sum notation $S \oplus T$ (sum with trivial intersection); unique decomposition $v = s + t$.
Coming soon in this course: $V = S \oplus S^\perp$ — projection (C10) splits any vector into a subspace part and an orthogonal part.

C6. Span, linear independence, basis, dimension in $\mathbb{R}^n$ / $\mathbb{C}^n$

Core theory

Span: all linear combinations. Independent: $\sum \alpha_i v_i = 0$ only for all $\alpha_i = 0$. Basis: independent and spanning. Dimension: size of any basis (well-defined).
Three instant counts — each is a complete one-line justification:
1. Fewer than $n$ vectors cannot span $\mathbb{R}^n$.
2. More than $n$ vectors in $\mathbb{R}^n$ cannot be independent.
3. Exactly $n$ independent vectors in $\mathbb{R}^n$ are automatically a basis (and vice versa: $n$ spanning vectors are automatically independent).
Mechanical test: set $\alpha_1 v_1 + \dots + \alpha_k v_k = 0$, read off one equation per coordinate, solve. A coordinate where only one vector is nonzero kills that coefficient immediately — scan for such rows first (e.g. HW2 3(a): row 2 forces $\alpha_1 = 0$ instantly, and the rest collapses).
Spot dependencies by inspection before computing: $v_3 = 3v_1$ (HW2 3(b)); coordinate-pattern sums like $y = u + w + x$ (Sp25 MT1.4(c) — check first coords: $1 = 1+0+0$, then verify the rest).
Membership: $x \in \operatorname{span}$ iff the linear system $\sum \alpha_i v_i = x$ is consistent.
Uniqueness ⇔ independence: representation in a spanning set is unique iff the set is independent. If dependent with dependency relation $\sum c_i v_i = 0$, generate infinitely many alternatives: (any solution) $+ \lambda \cdot$ (the relation).
Completing to a basis (Dis 2A Q1(c)): to extend an independent set to a basis of $\mathbb{R}^n$, append any vector outside the current span — for two vectors with third coordinate $0$, anything with a nonzero third coordinate works (e.g. $e_3$) — then close with the count: $n$ independent vectors ⇒ basis.
Proving two spans are equal (Dis 2A Q2): set equality = double containment, $\subseteq$ both ways. Show each generator of the left is a combination of the right's generators, and vice versa. E.g. $\operatorname{span}{v_1, \dots, v_n} = \operatorname{span}{\alpha v_1, v_2, \dots, v_n}$ for $\alpha \ne 0$: one direction is $\alpha v_1$ already in the old span; the other is $v_1 = \frac1\alpha(\alpha v_1)$. Nonzero scaling of generators never changes a span — and this double-containment template is how any set-equality proof is graded.

How it's been tested

Sp25 MT1.4 (40 pts): the full chain — ${u,v,w}$ independent; not a basis for $\mathbb{R}^5$ (count 1); ${u,v,w,x,y}$ not a basis ($y = u+w+x$); ${u,v,w,x,z}$ independent → basis (count 3).
HW2 Q3: independence in $\mathbb{R}^4$, basis verdicts, expressing $x$ two different ways in a dependent set.
Fa24 MT1.3, Sp26 E1.3(b): independence via orthogonality / via exhibiting a dependence.

Traps

"Independent" and "spanning" are separate claims — when asked "is it a basis," address both, or invoke the count shortcut by name.
To show not a basis: either a dimension count or an explicit vector outside the span / explicit dependency. Exhibit it.
Asked for an alternative linear combination: don't re-solve — add a multiple of the dependency relation you already found.

Rank and pivot columns (the matrix view of everything above).
Coordinates relative to a basis — a change of coordinates is the theme uniting C1 (Cartesian↔polar) with all of this.

C7. Function & polynomial spaces: independence beyond $\mathbb{R}^n$

Core theory

Functions are vectors; independence now means $\alpha_1 f_1(t) + \dots + \alpha_k f_k(t) = 0$ for all $t$ forces all $\alpha_i = 0$. Four techniques, in order of frequency:

Evaluate at chosen points. "$= 0$ for all $t$" implies "$=0$ at any points you like" — pick points that decouple the system. For ${1, \cos t, \sin t}$: $t = 0, \frac\pi2, \pi$ gives $\alpha+\beta = 0$, $\alpha + \gamma = 0$, $\alpha - \beta = 0$ → all zero. (Evaluation gives necessary conditions — enough to force coefficients to zero, which is exactly what independence needs.)
Differentiate / Taylor. For monomials ${1, t, \dots, t^n}$: $p^{(k)}(0) = k!,a_k = 0$ for each $k$.
Root counting. A nonzero polynomial of degree $\le n$ has at most $n$ roots; one vanishing for all $t$ has infinitely many — contradiction, so it's the zero polynomial, so all coefficients vanish. (Both this and #2 are on the facts sheet as usable results.)
Parity & orthogonality. Nonzero mutually orthogonal vectors are independent (see C10). And an identity like an even function claimed equal to a combination of odd ones dies by symmetry.

Identities are dependency relations. $\sin^2 + \cos^2 = 1$ means $1\cdot\psi_0 - \psi_1 - \psi_2 = 0$: the set ${1, \sin^2(\frac{\pi}{2}t), \cos^2(\frac{\pi}{2}t)}$ is dependent (Fa25 E1.1(b)) even though every pair inside it is independent. Pairwise independence ≠ independence.

Polynomial spaces. $P_n$ (degree $\le n$) has dimension $n+1$, basis ${1, t, \dots, t^n}$, and the coefficient vector $p \in \mathbb{R}^{n+1}$ is the coordinate representation — polynomial questions become $\mathbb{R}^{n+1}$ questions. Dis 1B Q1(b) makes this precise: $p(t) \leftrightarrow (p_0, \dots, p_n)$ is a unique representation in both directions (a coordinate isomorphism), and each independent homogeneous constraint drops the dimension by one — $V = {p : p(0) = 0}$ forces $p_0 = 0$, leaving unique coordinates in $\mathbb{R}^n$. HW1 Q6(c) adds: the set of derivatives of $P_3$ polynomials is itself a vector space (it is exactly $P_2$) — linear images of vector spaces stay vector spaces. For transformed families like $A = {1+t, (1+t)^2}$:

Expand into standard coordinates: $\alpha(1+t) + \beta(1+t)^2 = (\alpha+\beta) + (\alpha + 2\beta)t + \beta t^2$.
Find structure shared by the whole span: every element vanishes at $t = -1$ (common root).
Not a basis for $P_2$: $\dim(\operatorname{span} A) = 2 < 3$, and the shared-root gives a clean excluded vector — the constant $1$ has $1(-1) = 1 \ne 0$, so $1 \notin \operatorname{span}(A)$.

Reading coefficients off graphs for $\operatorname{span}{1, \cos t}$: in $f = \alpha + \beta\cos t$, $\alpha$ = vertical midline = $\frac{\max + \min}{2}$, $\beta$ = amplitude = $\frac{\max - \min}{2}$ (sign from whether peaks sit at $t=0$). Membership vetoes: wrong parity (any $\lambda + \mu\cos t$ is even — an odd, nonzero function can't be in the span), wrong period, or failed point evaluations (three points → inconsistent system).

How it's been tested

Fa24 MT1.4: monomial independence (15 pts) + polynomial subspace (20 pts).
Sp25 MT1.2: full ${1+t, (1+t)^2}$ workup. Sp25 MT1.3: trig independence + graph reading + span membership.
Fa25 E1.1: the $\sin^2/\cos^2$ trap + orthogonality-implies-independence with an integral inner product.
Sp26 E1.3(b): exhibit the dependence $z = \frac{1}{\sqrt2}(x + y)$ among three signals.

Traps

The Pythagorean-identity ambush — before grinding, ask: do these functions satisfy an identity?
The zero vector is the zero function: "$f(t) = 0$ for all $t$," not "$f$ has a zero somewhere."
Justify evaluation points: you're choosing them; the conclusion "all coefficients are 0" must follow from the resulting system, so show the system.

Taylor coefficients as coordinates — $C^\infty$ functions echo the $P_n$ story.
Fourier series: the independence (in fact orthogonality) of ${1, \cos t, \sin t, \cos 2t, \dots}$ is its foundation. This exam's trig problems are Fourier's opening act.

C8. Inner products: four flavors, one conjugation rule

Core theory

All four definitions, one table (cheat-sheet material):

Space	Inner product
$\mathbb{R}^n$	$\langle x, y\rangle = x^\top y = \sum_k x_k y_k$
$\mathbb{C}^n$	$\langle x, y\rangle = x^\top y^* = \sum_k x_k y_k^$ — conjugate on the second argument* (this course's convention; conventions vary elsewhere)
DT signals	$\langle f, g\rangle = \sum_{n=-\infty}^{\infty} f[n],g[n]$ (conjugate second if complex)
CT signals on $[a,b]$	$\langle f, g\rangle = \int_a^b f(t),g(t),dt$

Properties (these power every proof in C11):

Linear in the first argument: $\langle \alpha x + \beta y, z\rangle = \alpha\langle x,z\rangle + \beta\langle y,z\rangle$.
Conjugate symmetry: $\langle y, x\rangle = \langle x, y\rangle^*$ (plain symmetry in real spaces). Consequence: in expansions of $|x \pm y|^2$, the cross terms are $\langle x,y\rangle + \langle y,x\rangle = 2\operatorname{Re}\langle x,y\rangle$.
Positive definiteness: $\langle x, x\rangle = |x|^2 \ge 0$, with equality iff $x = 0$. For signals: $E_z = \sum |z[n]|^2 = 0$ forces every term of a sum of nonnegatives to be zero, so $z[n] = 0$ for all $n$ — the unique zero-energy signal is the zero signal (Sp26 E1.2(a) verbatim).
The axioms are themselves gradeable content (Dis 1A Q3, HW1 Q3): prove symmetry, linearity, non-negativity, and the expansion $\langle x+y, x+y\rangle = \langle x,x\rangle + 2\langle x,y\rangle + \langle y,y\rangle$ from the coordinate definition, componentwise — cite the definition, never the property as its own proof. Same drill for conjugate identities (HW1 Q1): $z + z^* = 2\operatorname{Re}(z)$; $zz^* \ge 0$ with equality iff $z = 0$; $z$ real $\iff z = z^*$; $(zv)^* = z^* v^*$ — each is two lines of $a + ib$ algebra, and the iffs need both directions.

Computation tactics:

Parity shortcut: $\int_{-1}^{1}(\text{odd}) , dt = 0$. Even × odd = odd. This killed Fa25 E1.1(c) in one line: $\sin^2(\frac{\pi}{2}t)$ is even, $\sin(\pi t)$ is odd → $\langle\psi_1,\psi_3\rangle = 0$ → $\theta_{13} = \frac{\pi}{2}$ → orthogonal → independent.
Support overlap: for piecewise signals, integrate only where both are nonzero — the smaller support. Fa25 E1.3(c): $\psi_k$ lives on $[0, 2^{-k})$, $\psi_\ell$ on $[0,2^{-\ell}) \subseteq [0,2^{-k})$ for $k \le \ell$, so $\langle \psi_k, \psi_\ell\rangle = \sqrt{2^k}\sqrt{2^\ell}\cdot 2^{-\ell} = 2^{(k-\ell)/2}$. (Check: $k = \ell$ gives $1 = |\psi_k|^2$ ✓.)
Complex vectors: $\langle x, \mathbb{1}\rangle = \sum_n x[n]\cdot 1^* = \sum_n x[n]$ — for the 8th-roots vector this is $0$ (C3!), so $x \perp \mathbb{1}$ and $\theta = \frac{\pi}{2}$ (Sp26 E1.1(d)). Old concept, new clothing.

How it's been tested

Everywhere: Fa24 MT1.3(a) (orthogonality of complex vectors — the conjugation actually matters: $\langle\varphi_1,\varphi_3\rangle = \sum \varphi_1[k]\varphi_3[k]^*$), Fa25 E1.1(c), E1.3(c), Sp26 E1.1(d), E1.2(a).

Traps

Dropping the conjugate in $\mathbb{C}^n$. It silently didn't matter for the all-ones vector; it will matter the moment the second argument is genuinely complex.
$\langle x, y \rangle \ne \langle y, x\rangle$ in complex spaces — they're conjugates.
Integrating a product over the wrong interval (full domain instead of the overlap).
Claiming $E_z = 0 \Rightarrow z = 0$ without the "sum of nonnegative terms" sentence — that sentence is the proof.

Energy and correlation: $\langle f, g\rangle$ measures alignment/similarity of signals — the seed of matched filtering and correlation detection later in EECS.
Weighted inner products $\langle x, y\rangle_W$ — same axioms, different geometry.

C9. Norms: $\ell^1$, $\ell^2$, $\ell^\infty$ across vectors, DT, and CT

Core theory

The nine-cell table (cheat-sheet material):

	$\ell^1$	$\ell^2$	$\ell^\infty$
vector $x \in \mathbb{C}^n$	$\sum_k \lvert x_k\rvert$	$\sqrt{\sum_k \lvert x_k\rvert^2}$	$\max_k \lvert x_k\rvert$
DT signal	$\sum_n \lvert x[n]\rvert$	$\sqrt{\sum_n \lvert x[n]\rvert^2}$	$\sup_n \lvert x[n]\rvert$
CT signal	$\int \lvert x(t)\rvert,dt$	$\sqrt{\int \lvert x(t)\rvert^2 dt}$	peak value $\sup_t \lvert x(t)\rvert$

Norm axioms: positive definiteness ($|x| = 0 \iff x = 0$), absolute homogeneity ($|\alpha x| = |\alpha|,|x|$), triangle inequality. A norm need not come from an inner product — Fa24 MT1.5 is built entirely on respecting that.
Energy: $E_x = |x|_2^2 = \langle x, x\rangle$. The $\ell^2$ norm is the one induced by the inner product; it's the only one of the three satisfying the parallelogram law.
Closed forms via the given sum formulas. For $x = (1, 2, \dots, n)$: $|x|_\infty = n$, $|x|_1 = \frac{n(n+1)}{2}$, $|x|_2 = \sqrt{\frac{n(n+1)(2n+1)}{6}}$.
Complex entries use moduli. Every entry of the DFT-style vectors has $|\cdot| = 1$, so each $\varphi_k \in \mathbb{C}^4$ has $|\varphi_k|_2 = 2$ (one computation + "same moduli" covers all four); the 8th-roots vector has $|x|_2 = \sqrt8 = 2\sqrt2$, $|x|_1 = 8$ (no cancellation — moduli, not values!), $|x|_\infty = 1$.
Time-scaling effects (CT). Compress $x(t) \to x(2^n t)$: width shrinks by $2^n$; $\ell^\infty$ unchanged; $\ell^1$ shrinks by $2^{-n}$; $\ell^2$ shrinks by $2^{-n/2}$. That's why Fa25's family $\psi_n(t) = \sqrt{2^n},\psi_0(2^n t)$ carries amplitude $\sqrt{2^n}$: it's the exact factor making $|\psi_n|_2 = 1$ — unit-energy normalization, a design choice, not a coincidence. Results: $|\psi_n|_1 = 2^{-n/2}$, $|\psi_n|_2 = 1$, $|\psi_n|_\infty = 2^{n/2}$ — three norms, three different stories about the same "size."

How it's been tested

Fa24 MT1.1(a), MT1.3(b); Fa25 E1.3(b) (all three norms of $\psi_n$ as functions of $n$); Sp26 E1.1(e).

Traps

Leaving a $\sum$ or $\int$ in a final answer when a closed form was demanded.
$|x|_1$ of a complex vector: sum of $|x_k|$, never of raw entries (the roots-of-unity entries sum to 0; their moduli sum to 8).
Both height and width change under $\sqrt{2^n}\psi_0(2^n t)$ — plot accordingly, label both.
$\ell^\infty$ of a CT signal is the peak over the whole domain, not an endpoint value.

Unit balls: diamond ($\ell^1$), disk ($\ell^2$), square ($\ell^\infty$) — the fastest intuition for why they measure different things (sparsity vs. energy vs. peak).
The $\psi_n$ family is a Haar wavelet preview; unit-energy normalization is ubiquitous in signal processing and quantum mechanics alike.

C10. Orthogonality, angles, orthogonal expansion, projection

Core theory

Definitions: $x \perp y \iff \langle x, y\rangle = 0$. Angle: $\cos\theta = \frac{\langle x, y\rangle}{|x||y|}$ (real spaces; on the facts sheet). Common values: $\cos\theta = \frac{1}{\sqrt2} \Rightarrow \theta = \frac\pi4$; $\langle x,y\rangle = 0 \Rightarrow \theta = \frac\pi2$.
Constructing an orthogonal vector (Fa24 MT1.1(b)): one homogeneous equation $\langle y, y_\perp\rangle = 0$ in $n$ unknowns → $n-1$ degrees of freedom. Fast recipe: zero out all but two coordinates, then swap those two and negate one: $y = (1,2,3) \to y_\perp = (2,-1,0)$; $z = (1,2,3,4) \to z_\perp = (2,-1,0,0)$. State the check.
Orthogonal ⇒ independent: if nonzero $v_1, \dots, v_k$ are mutually orthogonal and $\sum \alpha_i v_i = 0$, inner-product both sides with $v_j$: everything dies except $\alpha_j |v_j|^2 = 0$, so $\alpha_j = 0$. (This is why Fa25 E1.1(c)'s orthogonality conclusion instantly gives independence.)
Expansion in an orthogonal basis — no system solving. If ${\varphi_k}$ are mutually orthogonal and $v = \sum_k \alpha_k \varphi_k$, then inner-producting with $\varphi_k$ isolates $$\alpha_k = \frac{\langle v, \varphi_k\rangle}{|\varphi_k|^2}.$$ Fa24 MT1.3(c): $e_1 = \sum \alpha_k \varphi_k$ with $\langle e_1, \varphi_k\rangle = (\varphi_k[0])^* = 1$ and $|\varphi_k|^2 = 4$, so $\alpha_k = \frac14$ for all $k$ — and indeed $\frac14(\varphi_0+\varphi_1+\varphi_2+\varphi_3) = e_1$ (all other coordinates are 4-term roots-of-unity sums = 0; C3 again). This formula is the DFT.
Gram matrix, orthogonal vs orthonormal (Dis 2B Q1): $G_{k\ell} = \langle \varphi_k, \varphi_\ell\rangle$; stacking the vectors as columns of $\Phi$ gives $G = \Phi^\top \Phi^*$. Orthogonal set ⟺ $G$ diagonal; orthonormal ⟺ $G = I$. The DFT vectors give $G = 4I$: orthogonal, not orthonormal — normalize by dividing each by its norm ($\varphi_k/2$), after which the coefficients are pure inner products. One matrix computation certifies a whole basis at once.
Worked (Dis 2B Q1(c)): $u = (1, 0, 3, 0)^\top$ in the ${\varphi_k}$ basis: $\alpha_\ell = \frac{\langle u, \varphi_\ell\rangle}{4}$ gives $\alpha_0 = 1$, $\alpha_1 = -\frac12$, $\alpha_2 = 1$, $\alpha_3 = -\frac12$. And $\operatorname{proj}_{\varphi_2}(u) = \frac{\langle u, \varphi_2\rangle}{|\varphi_2|^2}\varphi_2 = \alpha_2\varphi_2$ — projecting onto one basis vector's span is extracting that component of the expansion. Dis 2B Q3 runs the same machinery on 4-sample audio signals (mutual orthogonality check, then project the ensemble recording onto a voice sample).
Projection (HW2 Q4 — geometric derivation is the point):
1. Length: $|\operatorname{proj}_v(u)| = |u|\cos\theta$ (right-triangle trig on the figure).
2. Direction: unit vector $\frac{v}{|v|}$, so $\operatorname{proj}_v(u) = |u|\cos\theta \cdot \frac{v}{|v|}$.
3. Eliminate $\theta$ with $\cos\theta = \frac{\langle u, v\rangle}{|u||v|}$: $$\operatorname{proj}_v(u) = \frac{\langle u, v\rangle}{\langle v, v\rangle},v.$$ The error $e = u - \operatorname{proj}_v(u)$ is orthogonal to $v$, and $|e|$ is the minimum distance from $u$ to the line through $v$ — projection = best approximation. (Note $\alpha_k$ above is exactly a projection coefficient: expansion in an orthogonal basis is projection onto each axis.)
Distance geometry / law of cosines: $|x - y|^2 = |x|^2 + |y|^2 - 2\langle x, y\rangle$. HW2 Q6(b): $x = (-3,3)$, $y = (4,4)$: $\langle x,y\rangle = 0$, so distance $= \sqrt{18 + 32} = 5\sqrt2$ — spot the orthogonality, skip the grind.
Nearest-neighbor classification (Sp25 MT1.1): "closer to $a$ than to $b$" $\iff |x-a|^2 \le |x-b|^2 \iff$ (expand, $|x|^2$ cancels) a linear inequality — the boundary is the perpendicular bisector of segment $ab$. Three prototypes → three bisectors meeting at the circumcenter; for an equilateral triangle it sits $\frac13$ up each bisector. Same fact as C2's $|z-a| = |z-b|$ locus.

How it's been tested

Fa24 MT1.1(b), MT1.3 (25 pts of orthogonal machinery); Sp25 MT1.1 (25 pts); Fa25 E1.1(c); Sp26 E1.1(d), E1.3(c); HW2 Q4, Q6.

Traps

Orthogonal basis ≠ orthonormal: divide by $|\varphi_k|^2$ (here 4). Forgetting the normalization is the classic error.
In $\mathbb{C}^n$, compute $\langle v, \varphi_k\rangle$ with the course's conjugation convention — conjugate lands on $\varphi_k$.
Perpendicular-bisector shading: verify with the prototype itself (it must lie in its own region).
The projection formula's denominator is $\langle v, v\rangle$, not $|v|$.

Least squares — projection onto a subspace spanned by many vectors; this course is heading straight there.
Gram–Schmidt: manufacturing orthogonal bases via repeated projection-and-subtract.
Classification/clustering as pure geometry — Sp25 MT1.1 is a 1-nearest-neighbor classifier, an actual ML algorithm.

C11. The inequality & identity toolkit (where the proof points live)

Core theory

The master move, behind nearly every proof problem on all four papers: convert norm statements to inner products, expand bilinearly, then cancel or bound. $$|x \pm y|^2 = \langle x \pm y,, x \pm y\rangle = |x|^2 \pm 2\operatorname{Re}\langle x, y\rangle + |y|^2 \quad (\text{real case: } \pm 2\langle x,y\rangle).$$

The toolkit built from it:

Cauchy–Schwarz: $|\langle x, y\rangle| \le |x||y|$, equality iff $x, y$ are colinear (linearly dependent). $\mathbb{R}^2$ proof (HW2 Q5): write $v = r(\cos\theta, \sin\theta)$, $w = t(\cos\varphi, \sin\varphi)$; then $|v| = r$, $|w| = t$ (Pythagorean identity), $\langle v, w\rangle = rt\cos(\theta - \varphi)$ (angle-difference identity), and $-1 \le \cos(\theta-\varphi) \le 1$ delivers both bounds. Equality ⟺ $\theta - \varphi \in {0, \pi}$ ⟺ same or opposite direction.
Triangle inequality from CS (HW2 Q6(c)): $|x+y|^2 = |x|^2 + 2\langle x,y\rangle + |y|^2 \le |x|^2 + 2|x||y| + |y|^2 = (|x| + |y|)^2$, take square roots.
Reverse triangle inequality (Fa24 MT1.5): $-|x-y| \le |x| - |y| \le |x - y|$. Proof uses only norm axioms — the add-and-subtract trick: $|x| = |(x-y) + y| \le |x-y| + |y|$, rearrange; swap roles of $x, y$ for the other side. Read the restrictions: the problem forbids assuming $\mathbb{R}^n$, a specific norm, or any inner product — using them caps you at 80%. Know which tools each proof is licensed to use.
Parallelogram law (Fa25 E1.2), valid in any inner-product space: $|x+y|^2 + |x-y|^2 = 2|x|^2 + 2|y|^2$ — expand both with the master move; the $\pm 2\operatorname{Re}\langle x,y\rangle$ cross terms cancel. (Contrast with the reverse triangle inequality: that one holds for every norm; this one is the fingerprint of inner-product norms. The two problems are a matched pair about which structure you're standing on.)
Iff-identities (Sp26 E1.2(c),(d), real spaces) — prove both directions; a chain of equivalences does both at once:
- $|x+y| = |x-y| \iff \langle x, y\rangle = 0$ (expand both sides; equality ⟺ $4\langle x,y\rangle = 0$).
- $\langle x+y,, x-y\rangle = |x|^2 - |y|^2$, so it's zero $\iff |x| = |y|$ (uses symmetry $\langle x,y\rangle = \langle y,x\rangle$ — flag that this cancellation is a real-space privilege).
CS applications — AM–GM (Sp25 MT1.5): the "judicious vector" trick. $n=2$: $u = (\sqrt x, \sqrt y)$, $v = (\sqrt y, \sqrt x)$ gives $\langle u,v\rangle = 2\sqrt{xy} \le |u||v| = x + y$. $n=3$: cyclic shift $a = (u,v,w)$, $b = (v,w,u)$ + the provided algebraic identity + substitutions $u = \sqrt[3]{x}$ etc. Payoff problem: among rectangles of fixed perimeter, the square maximizes area — AM–GM's equality case ($x = y$).
CS applications — norm comparison & numeric bounds (Dis 1B Q2): $|v|_1 \le \sqrt{n},|v|_2$ — for $v_i \ge 0$, pair $v$ with the all-ones vector: $|v|_1 = \langle v, \mathbb{1}\rangle \le |v|_2 |\mathbb{1}|_2 = \sqrt{n},|v|_2$ (in general, run it on the entrywise-modulus vector, whose $\ell^2$ norm equals $|v|_2$). Numeric flavor: $1(100) + 2(99) + \cdots + 100(1) \le 1^2 + \cdots + 100^2$ is CS on $x = (1, \dots, 100)$, $y = (100, \dots, 1)$ — $y$ is a permutation of $x$, so $|x||y| = |x|^2$. The transferable skill: to bound a sum, ask whose inner product is this?

Proof craft (graded explicitly):

Open by stating what must be shown; close by saying it's shown.
"Show" / "prove" ⇒ argument from definitions, not an example. "Determine whether" ⇒ verdict + proof or explicit counterexample.
Both directions for every iff.
You may cite an earlier part's result without having proven it.
Respect closed-form instructions; respect "you may not assume …" restrictions to the letter.

How it's been tested

Fa24 MT1.5 (20 pts), Sp25 MT1.5 (30 pts), Fa25 E1.2 (25 pts), Sp26 E1.2(c),(d) (32 pts), HW2 Q5–Q6. An inequality/identity proof has appeared on every paper.

Traps

Using an inner product where only a norm is given (or a specific norm where a generic one is demanded).
Dropping $\operatorname{Re}(\cdot)$ in complex expansions; assuming $\langle x,y\rangle = \langle y,x\rangle$ in $\mathbb{C}^n$.
Proving one direction of an iff and stopping.
Quoting CS equality as "$\theta = 0$" only — antiparallel ($\theta = \pi$) also achieves $|\langle x,y\rangle| = |x||y|$.

Polarization identity: recover $\langle x, y\rangle$ purely from norms — the converse craft of the parallelogram law.
Minkowski's inequality = triangle inequality for general $\ell^p$; Hölder generalizes CS. These four inequalities are the workhorses of the entire signals/optimization sequence.

C12. Signals as vectors: DT/CT fluency

Core theory

Types: DT signal $x : \mathbb{Z} \to \mathbb{C}$ (a sequence; plot as stems); CT signal $x : \mathbb{R} \to \mathbb{C}$. Both form vector spaces under pointwise addition and scaling — which is why every concept above applies verbatim to signals.
Transformations to plot on sight: $x(2^n t)$ compresses horizontally by $2^n$ (the support $[0,1)$ becomes $[0, 2^{-n})$); $c,x(t)$ scales amplitude. Combined in $\psi_n(t) = \sqrt{2^n},\psi_0(2^n t)$: taller by $\sqrt{2^n}$, narrower by $2^n$ — always draw and label both changes, with tick values on both axes.
DT complex exponentials: $x[n] = e^{i\omega n}$ is periodic iff $\frac{\omega}{2\pi}$ is rational; for $\omega = \frac\pi4$, period $N = 8$, and the samples are exactly the 8th roots of unity (C3). One period's samples assemble into a vector $x \in \mathbb{C}^8$ — the bridge move Sp26 E1.1 is built on: signal → vector → apply C8/C9/C10.
Vocabulary to parse set definitions instantly: support (where the signal is nonzero), bounded ($|x[n]| \le B_x < \infty$ for all $n$), energy ($E_x = |x|_2^2$), $\ell^2(\mathbb{Z})$ = finite-energy signals, $\ell^\infty(\mathbb{Z})$ = bounded signals. Fa25 E1.4 and Sp26 E1.2 are subspace problems (C4) wearing this vocabulary.
The recent papers are ~60–70% signal-framed. The skill being tested is translation: strip the signal costume, find the linear-algebra question underneath, answer it, translate back.

How it's been tested

Fa25 E1.3(a) (plot $\psi_1, \psi_2, \psi_n$ — 15 pts just for labeled plots); Sp26 E1.1 (68 pts riding on one complex exponential signal).

Traps

$x(2t)$ compresses; it does not stretch. Check with the support endpoint: $\psi_0(2t) \ne 0$ needs $0 \le 2t < 1$, i.e. $t \in [0, \frac12)$.
Unlabeled plots earn partial credit at best — label heights ($\sqrt2$, $2$, $\sqrt{2^n}$) and widths ($\frac12$, $\frac14$, $2^{-n}$).
On the unit circle, mark which point is $n = 0, 1, \dots, 7$ (start at $1$, step counterclockwise by $\frac\pi4$).

Kronecker delta & unit step (your lecture notes) are the canonical DT building blocks — expect them as raw material.
The $\psi_n$ family is the Haar wavelet system; sampling a CT exponential into $\mathbb{C}^N$ is the doorstep of the DFT.

C13. The concrete signal toolbox: special signals, parity, delta & step decompositions

Core theory

Named signals — plot and classify on sight (Dis 1A Q1):

Signal	Definition	Domain	Parity
Heaviside step $u(t)$	$0$ for $t<0$; $1$ for $t\ge0$	CT	neither
DT unit step $u[n]$	$0$ for $n<0$; $1$ for $n\ge0$	DT	neither
DT ramp $r[n]$	$0$ for $n<0$; $n$ for $n\ge0$	DT	neither
Rect $\operatorname{rect}(t)$	$1$ for $\lvert t\rvert \le \frac12$; else $0$	CT	even
Kronecker delta $\delta[n]$	$1$ at $n=0$; else $0$	DT	even

Parity: even ⟺ $x(-t) = x(t)$ (mirror across the vertical axis); odd ⟺ $x(-t) = -x(t)$ (180° rotation about the origin — forces $x(0) = 0$). Products: even·even = even, odd·odd = even, even·odd = odd — the engine behind C8's parity shortcut. Most signals are neither; that's a legitimate answer, and the step and ramp are the standard examples.
Sketch arithmetic (Dis 1A Q2): to draw $f + g$ or $3f - g$, work sample by sample over the stated window (stems for DT). This is vector addition made visible — signals combine coordinatewise because they are vectors.
Every DT signal is a combination of shifted deltas (HW1 Q7 — the most important equation in HW1): $$x[n] = \sum_{m=-\infty}^{\infty} x[m],\delta[n-m].$$ Read it correctly: $n$ is fixed, $m$ runs, and exactly one term survives ($m = n$). The samples $x[m]$ are the coordinates of the signal in the basis of shifted deltas ${\delta[n-m]}_{m \in \mathbb{Z}}$ — "any signal decomposes this way, uniquely" is a basis statement (C6 in signal clothing).
Delta ↔ step conversions — both directions:
- $\delta[n] = u[n] - u[n-1]$ (a step minus its one-sample delay leaves a single spike).
- $u[n] = \sum_{k=0}^{\infty} \delta[n-k]$ (a step is a train of deltas from $0$ onward).
Steps as building blocks (HW1 Q7(c),(d)): substitute the first conversion into the delta decomposition — for a signal that starts (zero before some time), $x[n] = \sum_m (x[m] - x[m-1]),u[n-m]$: the step coefficients are the jumps of the signal, and "fewest terms" means one term per jump. Raw infinite combinations admit redundancies — $u[n]$ itself also equals the telescoping sum $\sum_{k\ge0}(u[n-k] - u[n-k-1])$ — so uniqueness needs a restriction on which combinations count; once every signal has exactly one representation, the word for the set is basis, Q7(d)'s punchline.

How it's been tested

Dis 1A Q1–Q2: plot/classify the five named signals; sketch $f+g$ and $3f-g$.
HW1 Q7: delta decomposition of a plotted signal; $\delta \leftrightarrow u$ conversions; step decomposition with fewest terms; uniqueness → basis.
Dis 2B Q3: 4-sample signals as $\mathbb{C}^4$ vectors — orthogonality and projection on actual stem plots.
Fa25 E1.3's $\psi_0$ is a shifted/scaled rect, equivalently $u(t) - u(t-1)$: past exams already build from these blocks.

Traps

The ramp here is DT ($f : \mathbb{Z} \to \mathbb{R}$) — stems, not a line.
Their rect is closed at the edges: $\operatorname{rect}(\pm\frac12) = 1$.
Shift direction: $\delta[n-m]$ spikes at $n = m$ — for $m > 0$ that's a shift to the right. This exact confusion already surfaced once in your lecture notes; drill it until automatic.
In $x[m],\delta[n-m]$, the sample $x[m]$ is a number (the coefficient); $\delta[n-m]$ is the vector. Don't blur them.
Parity is checked against the definition at $-t$, not eyeballed from an asymmetric plotting window.

The delta decomposition is the sifting property — the identity that becomes convolution once systems enter the course.
HW1 Q7's "popular choice" of alternative basis is the Fourier basis: the Dis 2B / Fa24 MT1.3 vectors $\varphi_k$ are its 4-point version. C10 and C13 are the same story from both ends.
Even ⊕ odd: every signal splits uniquely as $x_e(t) = \frac{x(t)+x(-t)}{2}$ plus $x_o(t) = \frac{x(t)-x(-t)}{2}$ — the even and odd signals form subspaces with trivial intersection whose sum is the whole space: a live instance of C5's direct sum.

Coverage matrix

Paper	Problem	Concepts	Skill in one line
Fa24	MT1.1	C9, C10	closed-form norms of $(1..n)$; construct orthogonal vectors
Fa24	MT1.2	C1, C2, C3	Cartesian forms; $i^k$ sums; disk & bisector loci
Fa24	MT1.3	C3, C8, C9, C10	DFT-vector orthogonality, norms, orthogonal expansion
Fa24	MT1.4	C7, C4	monomial independence; polynomial-roots subspace
Fa24	MT1.5	C11	reverse triangle inequality, generic norm only
Sp25	MT1.1	C10, C2	nearest-neighbor regions via perpendicular bisectors
Sp25	MT1.2	C7, C6	span of ${1+t,(1+t)^2}$, common root, basis verdict
Sp25	MT1.3	C7	trig independence; graph → coefficients; span membership
Sp25	MT1.4	C6	independence/basis chain in $\mathbb{R}^5$
Sp25	MT1.5	C11	AM–GM via CS; fixed-perimeter optimization
Sp25	MT1.6	C1, C2, C3	$z=\lvert z\rvert$; $z^2=(z^*)^2$; $\sum_0^{999} i^k$
Fa25	E1.1	C7, C8	$\sin^2/\cos^2$ trap; orthogonality via parity integral
Fa25	E1.2	C11, C8	parallelogram law in $\mathbb{C}^n$
Fa25	E1.3	C12, C9, C8	plot scaled family; three norms of $\psi_n$; overlap inner product
Fa25	E1.4	C4	five subspace verdicts (60 pts)
Sp26	E1.1	C1, C3, C12, C8, C9	8th-roots signal: plot, sum, angle to $\mathbb{1}$, norms
Sp26	E1.2	C4, C8, C11	$\ell^2$ subspace; zero-energy signal; two iff-identities
Sp26	E1.3	C8, C9, C7, C10	CT inner products: unit norms, dependence, angles
HW2	Q1	C1, C4	complex sets as $\mathbb{R}^2$ subspaces
HW2	Q2	C5	$S+T$, $S\cap T$, basis union, dimension formula
HW2	Q3	C6	mechanical independence/basis; non-unique representation
HW2	Q4	C10	geometric derivation of the projection formula
HW2	Q5	C11	CS proof in $\mathbb{R}^2$ via polar coordinates
HW2	Q6	C9, C10, C11	distances; triangle inequality from CS
HW1	Q1	C2	conjugate-identity proofs (iffs, both directions)
HW1	Q2	C8, C9	$e_i$/selector products; $\mathbb{1}^\top x$ = sum of entries; $x^\top x = \\|x\\|^2$
HW1	Q3	C8	prove the inner-product axioms in $\mathbb{R}^n$
HW1	Q4	C4	parametrized-set subspace verdicts (absorbable vs genuine offsets)
HW1	Q5	C10	perpendicular ⇒ zero inner product, via slopes
HW1	Q6	C7, C4	$P_3$ as a vector space; monomial basis; derivative image
HW1	Q7	C13, C6	delta/step decompositions; uniqueness ⇒ basis
Dis 1A	Q1–3	C13, C8	special signals & parity; sketch arithmetic; axiom proofs
Dis 1B	Q1	C7, C4	$P_n \leftrightarrow \mathbb{R}^{n+1}$ coordinates; root-constraint subspace
Dis 1B	Q2	C11	numeric CS bound; $\\|v\\|_1 \le \sqrt n\\|v\\|_2$
Dis 2A	Q1–3	C6	span membership; completing a basis; $k$ vs $n$ trichotomy; span-equality proof
Dis 2B	Q1	C10, C8	Gram matrix; orthonormalization; coefficient formula; projection
Dis 2B	Q2	C1	inverse Euler; De Moivre
Dis 2B	Q3	C13, C10	audio signals as $\mathbb{C}^4$ vectors; orthogonality; projection

Cheat-sheet checklist

The facts page already gives you CS, triangle, geometric sum, angle formula, $\sum k$, $\sum k^2$, polynomial facts, trig values — don't waste sheet space on those. Spend it on:

The subspace verdict catalog (C4 table) + the quantifier warning.
The four inner-product definitions with the conjugation convention (C8 table).
The nine-cell norm table + CT time-scaling effects on each norm (C9).
Expansion coefficients $\alpha_k = \frac{\langle v, \varphi_k\rangle}{|\varphi_k|^2}$ and projection $\operatorname{proj}_v(u) = \frac{\langle u,v\rangle}{\langle v,v\rangle}v$.
The master-move expansions of $|x\pm y|^2$ (real and complex versions) + parallelogram law + reverse-triangle trick $x = (x-y)+y$.
$i^k$ cycle rules, roots-of-unity sum = 0 (both proofs, one line each), $e^{i\pi/4} = \frac{\sqrt2}{2}(1+i)$.
Perpendicular-bisector facts; parity shortcut $\int_{-1}^1 \text{odd} = 0$.
CS/AM–GM equality conditions (colinear; $x=y$).
Inverse Euler: $\cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2}$, $\sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}$; De Moivre via $(e^{i\theta})^n$.
$x[n] = \sum_m x[m]\delta[n-m]$; $\delta[n] = u[n]-u[n-1]$; $u[n] = \sum_{k\ge0}\delta[n-k]$; step coefficients = jumps $x[m]-x[m-1]$.
Parity rules (e·e = e, o·o = e, e·o = o) + the parities of the five named signals.
$|v|_1 \le \sqrt n,|v|_2$ (CS with $\mathbb{1}$); Gram test: orthonormal ⟺ $G = \Phi^\top\Phi^* = I$; normalize via $\varphi_k / |\varphi_k|$.

Space	Inner product
\(\mathbb{R}^n\)	\(\langle x, y\rangle = x^\top y = \sum_k x_k y_k\)
\(\mathbb{C}^n\)	\(\langle x, y\rangle = x^\top y^* = \sum_k x_k y_k^\) — conjugate on the second argument* (this course's convention; conventions vary elsewhere)
DT signals	\(\langle f, g\rangle = \sum_{n=-\infty}^{\infty} f[n],g[n]\) (conjugate second if complex)
CT signals on \([a,b]\)	\(\langle f, g\rangle = \int_a^b f(t),g(t),dt\)

	\(\ell^1\)	\(\ell^2\)	\(\ell^\infty\)
vector \(x \in \mathbb{C}^n\)	\(\sum_k \lvert x_k\rvert\)	\(\sqrt{\sum_k \lvert x_k\rvert^2}\)	\(\max_k \lvert x_k\rvert\)
DT signal	\(\sum_n \lvert x[n]\rvert\)	\(\sqrt{\sum_n \lvert x[n]\rvert^2}\)	\(\sup_n \lvert x[n]\rvert\)
CT signal	\(\int \lvert x(t)\rvert,dt\)	\(\sqrt{\int \lvert x(t)\rvert^2 dt}\)	peak value \(\sup_t \lvert x(t)\rvert\)

Set	Verdict	Why
\({x \in \mathbb{R}^n : \frac1n\sum x_k = 0}\) (zero mean)	✅	homogeneous linear constraint
CT signals with \(x(t)=0\) for $	t	>1$ (limited support)
DT signals with \(x[0] = 1\)	❌	zero signal not in the set
Bounded signals \(\ell^\infty(\mathbb{Z})\)	✅	$
Finite-energy signals \(\ell^2(\mathbb{Z})\)	✅	addition closure via triangle ineq.: \(\\|x+y\\| \le \\|x\\|+\\|y\\| < \infty\)
Solutions of \(\frac{dy}{dt} + \alpha y = 0\), \(\alpha\) fixed	✅	linearity of \(\frac{d}{dt}\): \((y_1+y_2)' + \alpha(y_1+y_2) = 0\)
Same, but "for some \(\alpha\)" (union over all \(\alpha\))	❌	\(e^{-t} + e^{-2t}\) solves no single such ODE
Polynomials in \(P_n\) with \(v(0)=v(1)=0\)	✅	two homogeneous linear constraints
\({re^{i\pi/3} : r \in \mathbb{R}} \subset \mathbb{C}\cong\mathbb{R}^2\) (line through origin)	✅	1-D line; basis \({(\tfrac12, \tfrac{\sqrt3}{2})}\)
\({3e^{i\theta}}\) (circle radius 3)	❌	no zero; not closed under scaling
\({a + 2i : a \in \mathbb{R}}\) (shifted line)	❌	no zero
Purely imaginary axis \({bi}\)	✅	line through origin
\({(2(x+y),,x,,y) : x,y \in \mathbb{R}}\)	✅	a span in disguise: \(x(2,1,0)^\top + y(2,0,1)^\top\) — and every span is a subspace
\({(x,,y,,z+1) : x,y,z \in \mathbb{R}}\)	✅	the offset is absorbable: \(z+1\) sweeps all of \(\mathbb{R}\), so the set is \(\mathbb{R}^3\)
\({(x,,y,,x+1) : x,y \in \mathbb{R}}\)	❌	genuine inhomogeneous relation \(x_3 = x_1 + 1\); the zero vector fails it

Signal	Definition	Domain	Parity
Heaviside step \(u(t)\)	\(0\) for \(t<0\); \(1\) for \(t\ge0\)	CT	neither
DT unit step \(u[n]\)	\(0\) for \(n<0\); \(1\) for \(n\ge0\)	DT	neither
DT ramp \(r[n]\)	\(0\) for \(n<0\); \(n\) for \(n\ge0\)	DT	neither
Rect \(\operatorname{rect}(t)\)	\(1\) for \(\lvert t\rvert \le \frac12\); else \(0\)	CT	even
Kronecker delta \(\delta[n]\)	\(1\) at \(n=0\); else \(0\)	DT	even

EE 66 Exam 1 — Complete Concept Map (Fa26)

Exam profile (all four iterations)

Priority tiers for Fa26

C1. Complex numbers: Cartesian ↔ polar, Euler's formula

Core theory

How it's been tested

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C2. Conjugates, modulus, and loci in the complex plane

Core theory

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C3. Powers of \(i\), roots of unity, geometric sums

Core theory

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C4. Vector spaces and the subspace test

Core theory

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C5. Subspace algebra: \(S+T\), \(S \cap T\), dimension formula

Core theory

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C6. Span, linear independence, basis, dimension in \(\mathbb{R}^n\) / \(\mathbb{C}^n\)

Core theory

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C7. Function & polynomial spaces: independence beyond \(\mathbb{R}^n\)

Core theory

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C8. Inner products: four flavors, one conjugation rule

Core theory

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C9. Norms: \(\ell^1\), \(\ell^2\), \(\ell^\infty\) across vectors, DT, and CT

Core theory

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C10. Orthogonality, angles, orthogonal expansion, projection

Core theory

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C11. The inequality & identity toolkit (where the proof points live)

Core theory

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C12. Signals as vectors: DT/CT fluency

Core theory

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C13. The concrete signal toolbox: special signals, parity, delta & step decompositions

Core theory

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Coverage matrix

Cheat-sheet checklist