ECE513 - Vectorspaces and Signal Processing

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), …, ϕn(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 \(x\).

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