Linear Operators

An operator $T:X \Rightarrow Y$ is a mapping from $X$ to $Y$

$T$ is a linear operator iff

$T(\alpha x + \beta y) = \alpha T(x) + \beta T(y)$ for $x,y \in X, \alpha, \beta \in \mathbb{F}$

Range space of $T$

$R(T) = \{Tx | x \in X\}$

Null space of $T$

$N(T) = \{x \in X | Tx = 0\}$

Injectivity, Surjectivity and Bijectivity

Surjectivity (onto)

$R(T) = Y$

Injectivity (one-to-one)

$N(T) = \emptyset$

Bijectivity

$R(T) = Y$ and $N(T) = \emptyset$

Let $Z_k:l_2 \Rightarrow l_2$ be a linear transformation such that $Z_k x[n] = x[n - k]$

Is $Z_k$ bijective?

Yes, because the $Z_k$ has an inverse (shifting backward in time by $k$)

Is decimation surjective/injective/bijective?

Decimation is surjective, because any sequence can be formed by decimation, but the original signal cannot be reconstructed.

Left and Right inverses

$T^R:X \Rightarrow Y$ is a right inverse iff $TT^Ry = y$ for all $y \in R(T)$ $T^L:X \Rightarrow Y$ is a left inverse iff $T^LTx = x$ for all $x \in X$

Finite Dimensional LVS

Sampling and Interpolation

Polynomials of deg $\leq 7$ foo foo cachoo foo

Let $x ∈ span(ϕ(t), …, ϕ_n(t))

Sample $x(t)$ at $[t_1, ..., t_n]$

Recovery of a continuous $x$ (interpolation) can be formally written as

\[\hat{x} = argmin_{\hat{x}} \sum(x(t_i) - \hat{x}(t_i))^2\]

b-spline interpolation

$\mathcal{X} = span(\phi_1, .., \phi_n)$

$\phi_k = b_n(t - k)$ where n is the spline order

$b_0(t) = \begin{cases} 1, & |t| \leq 1/2 \\ 0, & else \end{cases}$
$b_1 = b_0 \ast b_0$
. . .
$b_n = b_0 \ast b_{n-1}$

$b_n$ is n-differentiable (t = 0).

$B^N[a,b]$ is finite dimensional, but $B^N[\mathbb{R}]$ is not

Coordinate Vector

coordinate vector

Let $B = [b_1, ..., b_n]$ be a basis of finite dimensional LVS $\mathcal{V}$. Let $v \in \mathcal{V}$ s.t. $v = x_1b_1 + ... + x_n b_n$.

Then $[v]_B = [x_1, ..., x_n]$ is called the coordinate vector.

Let $\mathcal{X} = span(1, t, t^2)$

$B_1 = [1, t, t^2]$, $B_2 = [1, 1 - t, t^2 - 4t + 2]$

Let $v = t^2 + 1$

then $[v]_{B_1} = [1\ 0\ 1]^T$,

Change of Basis

Suppose we have vectorspaces $\mathcal{X}$ and $\mathcal{Y}$ with bases $B_1$ and $B_2$, respectively.

Then $[\cdot]_{B_1}^{B_2}: \mathcal{X} \Rightarrow \mathcal{Y}$ is called a change of basis operator

Finding change of basis form linear transformation

Let $\mathcal{X} = P^3, \mathcal{Y} = P^2$, where $B_x = [1, t, t^2, t^3], B_y = [1, t, t^2]$

Suppose we have a linear transformation $D$ that takes the derivative of $x \in \mathcal{X}$, yielding $y \in \mathcal{Y}$

$x(t) = a_0 + a_1 t + a_2 t^2 + a_3 t^3$ $[x]_{B_x} = \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \end{bmatrix}$

$y(t) = a_1 + 2 a_2 t + 3 a_3 t^2$ $[y]_{B_y} = \begin{bmatrix} a_1 \\ 2 a_2 \\ 3 a_3 \end{bmatrix}$

Then the columns of $[D]_{B_x}^{B_y}$ are $[DB_{xi}]_{B_y}$, where $B_{xi}$ is the i-th column of $x$.

$[D]_{B_x}^{B_y} = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix}$