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# Singular value decomposition

## Problem

Let $\alpha = \{\bv_1, \bv_2, \bv_3\}$ with

$$
\bv_1 = \frac{1}{\sqrt{3}}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix},\
\bv_2 = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix},\
\bv_3 = \frac{1}{\sqrt{6}}\begin{bmatrix} 1 \\ 1 \\ -2 \end{bmatrix}.
$$

Let $\beta = \{\bu_1, \bu_2, \bu_3, \bu_4\}$ with

$$
\bu_1 = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \
\bu_2 = \frac{1}{\sqrt{2}}\begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \
\bu_3 = \frac{1}{\sqrt{2}}\begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \end{bmatrix}, \
\bu_4 = \frac{1}{\sqrt{2}}\begin{bmatrix} 0 \\ 0 \\ -1 \\ 1 \end{bmatrix}.
$$

Let $A$ be a $4\times 3$ matrix such that

$$
[f_A]_\alpha^\beta =
\begin{bmatrix}
3 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}.
$$

For

$$
\bx = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix},
$$

find $A\bx$.

## Thought

Recall that $[f_A]_\alpha^\beta [\bx]_\alpha = [A\bx]_\beta$. Therefore, we need to transform $\bx$ into $[\bx]_\alpha$ and then transform $[A\bx]_\beta$ into $A\bx$ to get the answer.

## Sample answer

Note that $\alpha$ is an orthonormal basis, so we have

$$
\bx = c_1\bv_1 + c_2\bv_2 + c_3\bv_3
$$

with $c_i = \inp{\bx}{\bv_i}$ for $i = 1,2,3$. By direct computation, we have

$$
\bx = \sqrt{3}\bv_1 - \frac{1}{\sqrt{2}}\bv_2 - \frac{3}{\sqrt{6}}\bv_3 \text{ and }
[\bx]_\alpha = \begin{bmatrix} \sqrt{3} \\ -\frac{1}{\sqrt{2}} \\ -\frac{3}{\sqrt{6}} \end{bmatrix}.
$$

Now we may compute

$$
[A\bx]_\beta = [f_A]_\alpha^\beta [\bx]_\alpha =
\begin{bmatrix} 3\sqrt{3} \\ -\sqrt{2} \\ 0 \\ 0 \end{bmatrix}.
$$

This means

$$
\begin{aligned}
A\bx &= 3\sqrt{3}\bu_1 -\sqrt{2}\bu_2 + 0\bu_3 + 0\bu_4 \\
&= \frac{3\sqrt{3}}{\sqrt{2}}\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} -
\begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix} =
\begin{bmatrix} \frac{3\sqrt{3}}{\sqrt{2}} + 1 \\ \frac{3\sqrt{3}}{\sqrt{2}} - 1 \\ 0 \\ 0 \end{bmatrix}.
\end{aligned}
$$

*This note can be found at Course website > Learning resources.*