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Expert-verified Found in: Page 411 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Find the singular values of ${\mathbf{A}}{\mathbf{=}}\left[\begin{array}{cc}1& 0\\ 0& -2\end{array}\right]$.

The singular values of A are ${\sigma }_{1}=2$ and ${\sigma }_{2}=1$

See the step by step solution

## Step 1 of 2: Given information

It is given that $\mathrm{A}=\left[\begin{array}{cc}1& 0\\ 0& -2\end{array}\right]$

By computing the eigenvalues of the square matrix, the singular values of A are found

${A}^{t}A=\left[\begin{array}{cc}1& 0\\ 0& -2\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& -2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 4\end{array}\right]$

Then, $\left(\mathrm{x}{|}_{2}-{\mathrm{A}}^{1}\mathrm{A}\right)=\left[\begin{array}{cc}\mathrm{x}-1& 0\\ 0& \mathrm{x}-4\end{array}\right]$

## Step 2 of 2: Find the singular value

Let the characteristic of polynomial ${\mathrm{A}}^{1}\mathrm{A}$ be $\mathrm{p}\left(\mathrm{x}\right)$

Then,

$p\left(x\right)=det\left(x{I}_{2}-{A}^{1}A\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(x-4\right)-0\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(x-4\right)$

Therefore, the characteristic polynomial is $\left(x-1\right)\left(x-4\right)$, which implies that the roots of $\mathrm{p}\left(\mathrm{x}\right)$ are 1,4 .

So, the eigenvalues of ${\mathrm{A}}^{\mathrm{t}}\mathrm{A}$ are ${\lambda }_{1}=4$ and ${\lambda }_{2}=1$

Thus the singular values of A are ${\sigma }_{1}=\sqrt{4}=2$ and ${\sigma }_{2}=\sqrt{1}=1$

Result: ${\sigma }_{1}=2$ and ${\sigma }_{2}=1$are the singular values of A. ### Want to see more solutions like these? 