## Subsection2.3.5SVD of nonsingular matrices

###### Homework2.3.5.1.

Let $A \in \Cmxm$ and $A = U \Sigma V^H$ be its SVD.

TRUE/FALSE: $A$ is nonsingular if and only if $\Sigma$ is nonsingular.

TRUE

Solution

$\Sigma = U^H A V \text{.}$ The product of square matrices is nonsingular if and only if each individual matrix is nonsingular. Since $U$ and $V$ are unitary, they are nonsingular.

###### Homework2.3.5.2.

Let $A \in \Cmxm$ and $A = U \Sigma V^H$ be its SVD with

\begin{equation*} \Sigma = \left( \begin{array}{ c c c c} \sigma_0 \amp 0 \amp \cdots \amp 0 \\ 0 \amp \sigma_1 \amp \cdots \amp 0 \\ \vdots \amp \vdots \amp \ddots \amp \vdots \\ 0 \amp 0 \amp \cdots \amp \sigma_{m-1} \end{array} \right) \end{equation*}

TRUE/FALSE: $A$ is nonsingular if and only if $\sigma_{m-1} \neq 0 \text{.}$

TRUE

Solution

By the last homework, $A$ is nonsingular if and only if $\Sigma$ is nonsingular. A diagonal matrix is nonsingular if and only if its diagonal elements are all nonzero. $\sigma_{0} \geq \cdots \geq \sigma_{m-1} \gt 0 \text{.}$ Hence the diagonal elements of $\Sigma$ are nonzero if and only if $\sigma_{m-1} \neq 0 \text{.}$

###### Homework2.3.5.3.

Let $A \in \Cmxm$ be nonsingular and $A = U \Sigma V^H$ be its SVD.

ALWAYS/SOMETIMES/NEVER: The SVD of $A^{-1}$ equals $V \Sigma^{-1} U^H \text{.}$

SOMETIMES

Explain it!

Solution

It would seem that the answer is ALWAYS: $A^{-1} = ( U \Sigma V^H )^{-1} = (V^H)^{-1} \Sigma^{-1} U^{-1} = V \Sigma^{-1} U^H$ with

\begin{equation*} \begin{array}{l} \Sigma^{-1} \\ ~~~=~~~~ \lt \gt \\ \left( \begin{array}{ c | c | c | c} \sigma_0 \amp 0 \amp \cdots \amp 0 \\ \hline 0 \amp \sigma_1 \amp \cdots \amp 0 \\ \hline \vdots \amp \vdots \amp \ddots \amp \vdots \\ \hline 0 \amp 0 \amp \cdots \amp \sigma_{m-1} \end{array} \right)^{-1} \\ ~~~=~~~~ \lt \gt \\ \left( \begin{array}{ c | c | c | c} 1/\sigma_0 \amp 0 \amp \cdots \amp 0 \\ \hline 0 \amp 1/\sigma_1 \amp \cdots \amp 0 \\ \hline \vdots \amp \vdots \amp \ddots \amp \vdots \\ \hline 0 \amp 0 \amp \cdots \amp 1/\sigma_{m-1} \end{array} \right). \end{array} \end{equation*}

However, the SVD requires the diagonal elements to be positive and ordered from largest to smallest.

So, only if $\sigma_0 = \sigma_1 = \cdots = \sigma_{m-1}$ is it the case that $V \Sigma^{-1} U^H$ is the SVD of $A^{-1} \text{.}$ In other words, when $\Sigma = \sigma_0 I \text{.}$

###### Homework2.3.5.4.

Let $A \in \Cmxm$ be nonsingular and

\begin{equation*} \begin{array}{rcl} A \amp = \amp U \Sigma V^H \\ \amp = \amp \left( \begin{array}{c | c | c } u_0 \amp \cdots \amp u_{m-1} \end{array} \right) \left( \begin{array}{c | c | c} \sigma_0 \amp \cdots \amp 0 \\ \hline \vdots \amp \ddots \amp \vdots \\ \hline 0 \amp \cdots \amp \sigma_{m-1} \end{array} \right) \left( \begin{array}{c | c | c } v_0 \amp \cdots \amp v_{m-1} \end{array} \right)^H \end{array} \end{equation*}

be its SVD.

The SVD of $A^{-1}$ is given by (indicate all correct answers):

1. $V \Sigma^{-1} U^H \text{.}$
2. $\displaystyle \left( \begin{array}{c | c | c } v_0 \amp \cdots \amp v_{m-1} \end{array} \right) \left( \begin{array}{c | c | c} 1/\sigma_{0} \amp \cdots \amp 0 \\ \hline \vdots \amp \ddots \amp \vdots \\ \hline 0 \amp \cdots \amp 1/\sigma_{m-1} \end{array} \right) \left( \begin{array}{c | c | c } u_0 \amp \cdots \amp u_{m-1} \end{array} \right)^H$
3. $\displaystyle \left( \begin{array}{c | c | c } v_{m-1} \amp \cdots \amp v_{0} \end{array} \right) \left( \begin{array}{c | c | c} 1/\sigma_{m-1} \amp \cdots \amp 0 \\ \hline \vdots \amp \ddots \amp \vdots \\ \hline 0 \amp \cdots \amp 1/\sigma_{0} \end{array} \right) \left( \begin{array}{c | c| c } u_{m-1} \amp \cdots \amp u_{0} \end{array} \right)^H .$
4. $( V P^H ) ( P \Sigma^{-1} P^H ) ( U P^H )^H$ where $P = \left( \begin{array}{c c c c} 0 \amp \cdots \amp 0 \amp 1 \\ 0 \amp \cdots \amp 1 \amp 0 \\ \vdots \amp \amp \vdots \amp \vdots \\ 1 \amp \cdots \amp 0 \amp 0 \end{array} \right)$

3. and 4.

Explain it!

Solution

This question is a bit tricky.

1. It is the case that $A^{-1} = V \Sigma^{-1} U^H \text{.}$ However, the diagonal elements of $\Sigma^{-1}$ are ordered from smallest to largest, and hence this is not its SVD.

2. This is just Answer 1. but with the columns of $U$ and $V \text{,}$ and the elements of $\Sigma \text{,}$ exposed.

3. This answer corrects the problems with the previous two answers: it reorders colums of $U$ and $V$ so that the diagonal elements of $\Sigma$ end up ordered from largest to smallest.

###### Homework2.3.5.5.

Let $A \in \Cmxm$ be nonsingular. TRUE/FALSE: $\| A^{-1} \|_2 = 1 / \min_{\| x \|_2 = 1} \| A x \|_2 \text{.}$

TRUE

Solution
\begin{equation*} \begin{array}{l} \| A^{-1} \|_2 \\ ~~~=~~~~ \lt {\rm definition} \gt \\ \max_{x \neq 0} \frac{\| A^{-1} x \|_2}{ \| x \|_2} \\ ~~~=~~~~ \lt {\rm algebra} \gt \\ \max_{x \neq 0} \frac{1}{\frac{\| x \|_2}{ \| A^{-1} x \|_2}} \\ ~~~=~~~~ \lt {\rm algebra~} \gt \\ \frac{1}{\min_{x \neq 0} \frac{\| x \|_2}{ \| A^{-1} x \|_2}} \\ ~~~=~~~~ \lt {\rm substitute~} z = A^{-1} x \gt \\ \frac{1}{\min_{A z \neq 0} \frac{\| A z \|_2}{ \| z \|_2}} \\ ~~~=~~~~ \lt A {\rm ~is~nonsingular} \gt \\ \frac{1}{\min_{z \neq 0} \frac{\| A z \|_2}{ \| z \|_2}} \\ ~~~=~~~~ \lt x = z/\| z \|_2 \gt \\ \frac{1}{\min_{\| x \|_2 =1} \| A x \|_2} \\ \end{array} \end{equation*}

In Subsection 2.3.2, we discussed the case where $A \in \R^{2 \times 2} \text{.}$ Letting $A = U \Sigma V^T$ and partitioning

\begin{equation*} A = \left( \begin{array}{c | c} u_0 \amp u_1 \end{array} \right) \left( \begin{array}{c | c} \sigma_0 \amp 0 \\ \hline 0 \amp \sigma_1 \end{array} \right) \left( \begin{array}{c | c} v_0 \amp v_1 \end{array} \right)^T \end{equation*}

yielded the pictures

$\R^{2} \text{:}$ Domain of $A \text{:}$

$\R^{2} \text{:}$ Range (codomain) of $A \text{:}$  This captures what the condition number $\kappa_2( A ) = \sigma_0 / \sigma_{n-1}$ captures: how elongated the oval that equals the image of the unit ball is. The more elongated, the greater the ratio $\sigma_0 / \sigma_{n-1} \text{,}$ and the worse the condition number of the matrix. In the limit, when $\sigma_{n-1} = 0 \text{,}$ the unit ball is mapped to a lower dimensional set, meaning that the transformation cannot be "undone."

###### Ponder This2.3.5.6.

For the 2D problem discussed in this unit, what would the image of the unit ball look like as $\kappa_2( A ) \rightarrow \infty \text{?}$ When is $\kappa_2( A ) = \infty \text{?}$