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### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Let be an orthogonal 2X2 matrix. Use the image of the unit circle to find the singular values of A.

The singular values of A are ${\mathrm{\sigma }}_{1}=1\mathrm{and}{\mathrm{\sigma }}_{2}=1$are derived using the theorem 8.3.2

See the step by step solution

## Step 1 of 2: Given information

It is given that A is an orthogonal 2x2 matrix.

## Step 2 of 2: Find the singular value

Let us have ${\mathrm{v}}_{1}=\left[\begin{array}{c}1\\ 0\end{array}\right]\mathrm{and}{\mathrm{v}}_{2}=\left[\begin{array}{c}0\\ 1\end{array}\right],\mathrm{where}\left\{{\mathrm{v}}_{1},{\mathrm{v}}_{2}\right\}$ forms an orthonormal basis of .

Here, the unit circle consists of the vectors of the form $\mathrm{x}=\mathrm{cos}\left(\mathrm{t}\right){\mathrm{v}}_{1}+\mathrm{sin}\left(\mathrm{t}\right){\mathrm{v}}_{1}$ and the image of the unit circle consists of the vectors of the form

$\mathrm{L}\left(\mathrm{x}\right)=\mathrm{cos}\left(\mathrm{t}\right)\mathrm{L}\left({\mathrm{v}}_{1}\right)+\mathrm{sin}\left(\mathrm{t}\right)\mathrm{L}\left({\mathrm{v}}_{2}\right)$

Therefore, the image is the ellipse whose semi-major and semi-minor axes are $\mathrm{L}\left({\mathrm{v}}_{1}\right)\mathrm{and}\mathrm{L}\left({\mathrm{v}}_{2}\right)$ and respectively.

The length of the axes are ${||\mathrm{L}\left({\mathrm{v}}_{1}\right)||}^{2}=\left({\mathrm{Av}}_{1}\right)\mathrm{x}\left({\mathrm{Av}}_{1}\right)={\mathrm{v}}_{1}^{\mathrm{T}}{\mathrm{A}}^{\mathrm{T}}{\mathrm{Av}}_{1}$

,$={\mathrm{v}}_{1}^{\mathrm{T}}{\mathrm{l}}_{2}{\mathrm{v}}_{1},\mathrm{where}\mathrm{A}\mathrm{is}\mathrm{an}\mathrm{orthogonal}\mathrm{matrix}\phantom{\rule{0ex}{0ex}}={\mathrm{v}}_{1}^{\mathrm{T}}{\mathrm{v}}_{1}=1\mathrm{as}{\mathrm{v}}_{1}\mathrm{is}\mathrm{an}\mathrm{unit}\mathrm{vector}\phantom{\rule{0ex}{0ex}}||\mathrm{L}{\left({\mathrm{v}}_{2}\right)}^{2}||=\left({\mathrm{Av}}_{2}\right)\mathrm{x}\left({\mathrm{Av}}_{2}\right)={\mathrm{v}}_{1}^{\mathrm{T}}{\mathrm{A}}^{\mathrm{T}}{\mathrm{Av}}_{2},\phantom{\rule{0ex}{0ex}}={\mathrm{v}}_{1}^{\mathrm{T}}{\mathrm{l}}_{2}{\mathrm{v}}_{1}\mathrm{where}\mathrm{A}\mathrm{is}\mathrm{an}\mathrm{orthogonal}\mathrm{matrix}\phantom{\rule{0ex}{0ex}}={\mathrm{v}}_{1}^{\mathrm{T}}{\mathrm{v}}_{1}=1\mathrm{as}{\mathrm{v}}_{2}\mathrm{is}\mathrm{an}\mathrm{unit}\mathrm{vector}$

Thus $||\mathrm{L}{\left({\mathrm{v}}_{2}\right)}^{2}||=1,||\mathrm{L}{\left({\mathrm{v}}_{2}\right)}^{2}||=1$

Thus, the singular values of A are ${\sigma }_{1}=1\mathrm{and}{\mathrm{\sigma }}_{2}=1$.

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