# Singular value decomposition of tricky matrix

Let $$A := \begin{bmatrix} -3 & 0 & 1 & 2 \\ 4 & 0 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ \end{bmatrix}$$

1. Compute a singular value decomposition (over $\mathbb{C}$) of $A$.

2. Compute the pseudo-inverse of $A$.

For (1), I have computed $$AA^T=\begin{matrix} 14 & -12 & 5 \\ -12 & 16 & 0 \\ 5 & 0 & 5 \\ \end{matrix}$$ Then to find the Eigenvalues, I computed $$\begin{matrix} 14-\lambda & -12 & 5 \\ -12 & 16-\lambda & 0 \\ 5 & 0 & 5-\lambda \\ \end{matrix}$$ and took the determinate of my new $3x3$ matrix and have now ended up with this $$=-\lambda^3+35\lambda^2-255\lambda+800$$ Now i'm stuck. I feel as though I am lost and/or have calculated wrong.

The last column of $AA^T$ should be $\left(\begin{array}{c} 5 \\ 0 \\ 5 \end{array}\right).$

Also, notice that the second column of $A$ is a zero column. the third column and the fourth column are multiple of each other. Hence the rank of the matrix is 2. Hence you are sure. One of the eigenvalue of $AA^T$ is $0$.

The characteristic equation that you should have obtained should have $0$ as one of the root. Your characteristic equation is wrong. The constant term has to be zero. Are you able to find the mistake?

Moore-Penrose Pseudoinverse matrix

$$\mathbf{A}^{\dagger} = \mathbf{V} \, \left[ \begin{array}{cc} \mathbf{S}^{-1} & \mathbf{0} \\ \mathbf{0} & 0 \end{array} \right] \, \mathbf{U}^{*} = % % \frac{1}{205} \left[ \begin{array}{rrr} -15 & 40 & 15 \\ 0 & 0 & 0 \\ 16 & 12 & 25 \\ 32 & 24 & 50 \\ \end{array} \right]$$

Construction

$$\mathbf{A}^{*} \mathbf{A} = \left[ \begin{array}{rrrr} 25 & 0 & -3 & -6 \\ 0 & 0 & 0 & 0 \\ -3 & 0 & 2 & 4 \\ -6 & 0 & 4 & 8 \\ \end{array} \right]$$

Characteristic polynomial: $$p(\lambda) = \lambda^{2} \left( \lambda^{2} - 35\lambda + 205 \right)$$

Eigenvalues: $$\lambda \left( \mathbf{A}^{*} \mathbf{A} \right) = \left( \frac{1}{2} \left( 35 \pm 9 \sqrt{5} \right),\, 0,\, 0 \right)$$ The matrix $\mathbf{A}$ has rank $\rho = 2$.

Singular values: $$\left( \sigma_{1}, \sigma_{2} \right) = \left( \sqrt{ \frac{1}{2} \left( 35 \pm 9 \sqrt{5} \right)} \right)$$ Define $$\mathbf{S} = \left[ \begin{array}{cc} \sigma_{1} & 0\\ 0 & \sigma_{2} \\ \end{array} \right], \quad % \Sigma = \left[ \begin{array}{cc} \mathbf{S} & \mathbf{0} \\ \mathbf{0} & 0 \\ \end{array} \right] = % \left( \begin{array}{cccc} \sigma_{1} & 0 & 0 & 0 \\ 0 & \sigma_{2} & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) %$$

Eigenvectors: $$% \color{blue}{v_{1}} = \color{blue}{\left[ \begin{array}{c} \frac{1}{4} \left(-3 \sqrt{5}-5\right) \\ 0 \\ \frac{1}{2} \\ 1 \\ \end{array} \right]}, \ % \color{blue}{v_{2}} = \color{blue}{\left[ \begin{array}{c} \frac{1}{4} \left(-3 \sqrt{5}+5\right) \\ 0 \\ \frac{1}{2} \\ 1 \\ \end{array} \right]}, \ % \color{red}{v_{3}} = \color{red}{\left[ \begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \\ \end{array}\right]}, \ % \color{red}{v_{4}} = \color{red}{\left[ \begin{array}{r} 0 \\ 0 \\ -2 \\ 1 \\ \end{array} \right]} %$$ $\color{blue}{Range\ space}$ vectors are in blue; $\color{red}{nullspace}$ in red. The normalized eigenvectors are the column vectors of domain matrix: $$\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}} = \left[ \begin{array}{cc} %% \sqrt{\frac{15}{2 \left(3 - \sqrt{5}\right)}} \color{blue}{\left[ \begin{array}{c} \frac{1}{4} \left(-3 \sqrt{5}-5\right) \\ 0 \\ \frac{1}{2} \\ 1 \end{array} \right]}, \, % \sqrt{\frac{15}{2 \left(3 + \sqrt{5}\right)}} \color{blue}{\left[ \begin{array}{c} \frac{1}{4} \left(-3 \sqrt{5}+5\right) \\ 0 \\ \frac{1}{2} \\ 1 \end{array} \right]} %% \end{array} \right].$$ The nullspace component vectors for the domain matrix: $$\color{red}{\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}} = \left[ \begin{array}{cc} %% \frac{1}{\sqrt{5}} \color{red}{\left[ \begin{array}{r} 0 \\ 0 \\ -2 \\ 1 \end{array} \right]}, \, % \color{red}{\left[ \begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array} \right]} %% \end{array} \right].$$

Given the singular values and the column vectors of the $\mathbf{V}$ matrix we can construct the range space component vectors for the domain matrix using $$\color{blue}{\mathbf{U}_{k}} = \frac{\mathbf{A} \color{blue}{\mathbf{V}_{k}}} {\sigma_{k}}$$ The column vectors are $$\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} = % \left[ \begin{array}{cc} %% \sqrt{\frac{15}{\left(35 + 9\sqrt{5}\right) \left(3 - \sqrt{5}\right)}} \color{blue}{\left[ \begin{array}{c} \frac{1}{4} \left(25 + 9\sqrt{5}\right) \\ -5-3\sqrt{5} \\ \frac{5}{2} \end{array} \right]}, \, % \sqrt{\frac{15}{\left(35 - 9\sqrt{5}\right) \left(3 - \sqrt{5}\right)}} \color{blue}{\left[ \begin{array}{c} \frac{1}{4} \left(25 - 9\sqrt{5}\right) \\ -5+3\sqrt{5} \\ \frac{5}{2} \end{array} \right]} %% \end{array} \right]. % % % %$$ The nullspace vectors are constructed using the orthogonalization process of Gram and Schmidt. For the codomain matrix they are: $$\color{red}{\mathbf{U}_{\mathcal{N}\left(\mathbf{A}^{*}\right)}} = \left[ \begin{array}{cc} %% \frac{1}{\sqrt{41}} \color{red}{\left[ \begin{array}{r} -4 \\ -3 \\ 4 \end{array} \right]} %% \end{array} \right].$$

The singular value decomposition is $$\mathbf{A} = \mathbf{U} \, \Sigma \, \mathbf{V}^{*} = % \left[ \begin{array}{cc} \color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} & \color{red} {\mathbf{U}_{\mathcal{N}\left(\mathbf{A}^{*}\right)}} \end{array} \right] % \Sigma \ % \left[ \begin{array}{cc} \color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}} & \color{red} {\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}} \end{array} \right]^{*} %$$

• Nice formatting! – Gabriel Romon Mar 12 '17 at 20:43