Exercise 6.3.5

Question

Find singular value decomposition of the matrix 
%%%%%%%%%%%%%%%%%
\begin{equation*}
    A = 
    \begin{bmatrix}
    2  &3  \
    0  &2
    \end{bmatrix}.
\end{equation*}
%%%%%%%%%%%%%%%%
Use it to find
\begin{enumerate}[label = \alph*)]
    \item $\max_{||\mathbf{x}|| \leqslant 1} ||A \mathbf{x}||$ and the vector where the maximum is attained;
    \item $\max_{||\mathbf{x}|| = 1} ||A \mathbf{x}||$ and the vector where the minimum is attained;
    \item the image $A(B)$ of the closed unit ball in $\mathbb{R}^2$, $B = \{\mathbf{x} \in \mathbb{R}^2 : ||\mathbf{x}|| \leqslant 1\}$. Describe $A(B)$ geometrically.
\end{enumerate}

Jing Huang · Accepted Answer

$
ewcommand{\T}{^\mathsf{T}}$
	extbf{Solution} (The SVD steps are ignored here.) \par
a) Suppose $A = W \Sigma V^*$, then $(A\mathbf{x}, A\mathbf{x}) = \mathbf{x}^* A^*A \mathbf{x} = \mathbf{x}^* V \Sigma^2 V^* \mathbf{x} = (V^* \mathbf{x})^* \Sigma^2 (V^* \mathbf{x})$. Define $\mathbf{y} = [y_1 \; y_2]\T = V^*\mathbf{x}$. Because $V$ is orthogonal, then $\mathbf{y}$ also lies in the unit ball. Thus
%%%%%%%%%%%%%%%%%
\begin{equation*}
\begin{split}
    (A\mathbf{x}, A\mathbf{x}) &= \mathbf{y}^* \Sigma^2 \mathbf{y}  \
    &=
    \begin{bmatrix}
    y_1  &y_2
    \end{bmatrix}
    \begin{bmatrix}
    16  &0  \
    0  &1
    \end{bmatrix}
    \begin{bmatrix}
    y_1  \
    y_2
    \end{bmatrix}  \
    &= 16 y_1^2 + y_2^2.
\end{split}
\end{equation*}
%%%%%%%%%%%%%%%%
$y_1^2 + y_2^2 \leqslant 1$. Hence the maximum is $16$ attained when $\mathbf{y} = [1 \; 0]\T$. Corresponding $\mathbf{x}$ can be solved by $\mathbf{x} = V \mathbf{y}$.  \par
b) Similarly, the minimum is $1$ attained when $\mathbf{y} = [0 \; 1]\T$.  \par
c) Ellipse.

Exercise 6.3.5

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Exercise 6.3.5

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