2 Vector Spaces and norms
useless information graveyard
\(\sum_{n=0}^{\infty} a^n = \frac{1}{1-a}\) where \(|a| < 1\)
Norms
1-norm for DT Signals
\(x: \mathbb{Z} \to \mathbb{C}\)
\(\|x\|_1 = \sum_{n=-\infty}^{\infty} |x[n]|\)
if \(\|x\|_1 < \infty \Rightarrow x \in \ell(\mathbb{Z})\)
\(\ell(\mathbb{Z}) = \{x: \mathbb{Z} \to \mathbb{C} \mid \|x\|_1 = \sum_{n=-\infty}^{\infty} |x[n]| < \infty\}\)
what is the 1-norm for this vector?
is \(\ell_1(\mathbb{Z})\) a subspace?
FIRST! is it a subspace of a superset?
it is a subspace of all complex discrete time signals \(\ell_1(\mathbb{Z}) \subset [\mathbb{Z} \to \mathbb{C}]\)
ok now we can check
- existence of a zero vector \(x[n] = 0\) for any \(n \in \mathbb{Z}\)
- closed under vector addition
- let \(x, y \in \ell_1(\mathbb{Z})\)
- \(\|x\|_1 < \infty\)
- \(\|y\|_1 < \infty\)
- let \(z = x + y\)
- \(\|z\|_1 = \|x + y\|_1 \leq \|x\|_1 + \|y\|_1 < \infty\)
- we know that each of these is finite, by assumption, so the sum is finite.
- that means that \(\|z\|_1\) is finite, and hence \(\ell_1\) is closed under vector addition.
- Closed on Scalar Multiplication
- let \(x \in \ell_1(\mathbb{Z})\)
- \(\|x\|_1 < \infty\)
- let \(z = cx\) for every \(c \in \mathbb{C}\)
- \(z[n] = c\,x[n]\) for all \(n \in \mathbb{Z}\)
- \(\|z\|_1\) defined as \(\sum_{n=-\infty}^{\infty} |y[n]| = \sum_{n=-\infty}^{\infty} |c\,x[n]| = \sum_n |c|\,|x[n]| = c \cdot \sum \ldots\) element of \(\ell_1\), hence closed, finish the proof urself
2 Norm
\(x = [x_1, \ldots, x_n] \in \mathbb{R}^n\) or \(\mathbb{C}^n\)
2 norm is the sum of the squares of these components to the power of 0.5
\(\|x\|_2 = (|x_1|^2 + \ldots + |x_n|^2)\)
what is the 2 norm of a DT signal is
we have signal \(x: \mathbb{Z} \to \mathbb{C}\)
\(\|x\|_2 = \sum_{n=-\infty}^{\infty} |x[n]|\)
you can stack entries for sandwiches or burgers. you forgot the one half, looks like the
energy is the square of the 2 norm of the function
just like \(\ell_1\) subspaces, we also have \(\ell_2\) spaces.
the set or space or whatever, and square sum of all signals.
square summable signals is denoted as square summable signals so that you don't geel the tuje of yh tg id giniy.r ig yhrtr id motr yhsn onr ddgnsl, er eoulf puy s trsl moniyot.
\(\ell_2(\mathbb{Z})\) is a subspace.
Infinity norm for signals
(we don't use maximum, we use supremum)
if \(x\) is a DT signal, the infinity norm of \(X\) is
\(\|x\|_\infty = \sup(\ldots, |x[-1]|. \text{ the upper limit within the set.})\)
the infinity norm is 1, we might still have it on the board. the very first non-zero lollipop from ur step function or something.
\(\|x\|_\infty = 1\)
\(\ell_\infty(\mathbb{Z}) \; \{x: \mathbb{Z} \to \mathbb{C}\}\)
we call this the set of bounded signals
we know that all the signal values are less than equal to \(B_x < \infty\)
\(|x[n]| < B_n < \infty\)
systems that take signals that are bounded, and you are assured the output of that system is bounded, those are special, we call them bounded input, bounded output systems.
Distance between vectors
what is a reasonable norm for fire station or emergency services?
the infinity norm, the maximum distance.
Inner Products!
when we lived in cartesian space, the dot product was the transpose of the first vector times the second vector
\(x \cdot y = x^T y = [x_1, \ldots, x_n][y_1, \ldots, y_n]^T = \sum_{k=0}^{n} x_k y_k\)
if \(x = [1 \; i]^T\)
\(x \cdot x = [1 \; i][1 \; i]^T = 1 + i^2 = 1 - 1 = 0\)
1 + i^2 =>we need something better!
\(V\): vector space \(x, y, z \in V\) \(\alpha, \beta \in \mathbb{C}\) \(\langle \cdot, \cdot \rangle: V \times V \to \mathbb{C}\)
Conjugate symmetry (Hermitian Property) \(\langle x, y \rangle = \langle y, x \rangle^*\) (\(\langle x, y \rangle\) is the complex conjugate of \(\langle y, x \rangle\))
\(\langle x, y + z \rangle = \langle x, y \rangle + \langle x, z \rangle\) Distributive Property
\(\langle ax, y \rangle = a\langle x, y \rangle\) Scaling Property
\(\langle x, x \rangle \geq 0\) w/ equality iff \(x = 0\) non-negativity (Positive Semi-definiteness)
Example: \(\langle x, by \rangle = b^* \langle x, y \rangle\) but why?
\(\langle x, by \rangle = \langle by, x \rangle^* = b^* \langle y, x \rangle^* = b^* \langle x, y \rangle\)
so scalars that show up in the first vector come out straight
if a scalar shows up in front of the second vector, it comes up complex conjugated.
let define an inner product in \(\mathbb{C}^n\)
\(x, y \in \mathbb{C}^n\)
\(\langle x, y \rangle = x^T y^*\) so the transpose of the first vector, times the complex conjugate of the second vector.
physicists would define it as \(x^{||} y = (x^*)^T y\)
in most engineering, it's the second object that's complex conjugated.
now according to this def, if we go back to our original example,
\(x = [1, i]\)
then \(\langle x, x \rangle = x^T x^* = (1 \cdot 1) + (i \cdot -i) = 1 + 1 = 2 = \|x\|_2^2\) (equal to the 2norm squared)
(1 * 1) + (i * -i) =>for time signals
\(x, y: \mathbb{Z} \to \mathbb{C}\)
\(\langle x, y \rangle = \sum_{n=-\infty}^{\infty} x[n]^T \, y[n]^*\)
for continuous time signals we integrate
\(\langle x, y \rangle = \int_{-\infty}^{\infty} x(t) \, y(t)^* \, dt\)