Question

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

Short answer

The singular values of A are ${\mathrm{\sigma }}_1=1\mathrm{and}{\mathrm{\sigma }}_2=1$are derived using the theorem 8.3.2

Step-by-step solution

of 2: Given information

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

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 $\mathrm{x}=\cos \left(\mathrm{t}\right){\mathrm{v}}_1+\sin \left(\mathrm{t}\right){\mathrm{v}}_1$and the image of the unit circle consists of the vectors of the form

$\mathrm{L}\left(\mathrm{x}\right)=\cos \left(\mathrm{t}\right)\mathrm{L}\left({\mathrm{v}}_1\right)+\sin \left(\mathrm{t}\right)\mathrm{L}\left({\mathrm{v}}_2\right)$

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 ${||\mathrm{L}\left({\mathrm{v}}_1\right)||}^2=\left({\mathrm{Av}}_1\right)\mathrm{x}\left({\mathrm{Av}}_1\right)={\mathrm{v}}_1^{\mathrm{T}}{\mathrm{A}}^{\mathrm{T}}{\mathrm{Av}}_1$

,$$\begin{gathered} ={\mathrm{v}}_1^{\mathrm{T}}{\mathrm{l}}_2{\mathrm{v}}_1,\mathrm{where}\mathrm{A}\mathrm{is}\mathrm{an}\mathrm{orthogonal}\mathrm{matrix} \\ ={\mathrm{v}}_1^{\mathrm{T}}{\mathrm{v}}_1=1\mathrm{as}{\mathrm{v}}_1\mathrm{is}\mathrm{an}\mathrm{unit}\mathrm{vector} \\ ||\mathrm{L}{\left({\mathrm{v}}_2\right)}^2||=\left({\mathrm{Av}}_2\right)\mathrm{x}\left({\mathrm{Av}}_2\right)={\mathrm{v}}_1^{\mathrm{T}}{\mathrm{A}}^{\mathrm{T}}{\mathrm{Av}}_2, \\ ={\mathrm{v}}_1^{\mathrm{T}}{\mathrm{l}}_2{\mathrm{v}}_1\mathrm{where}\mathrm{A}\mathrm{is}\mathrm{an}\mathrm{orthogonal}\mathrm{matrix} \\ ={\mathrm{v}}_1^{\mathrm{T}}{\mathrm{v}}_1=1\mathrm{as}{\mathrm{v}}_2\mathrm{is}\mathrm{an}\mathrm{unit}\mathrm{vector} \end{gathered}$$

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${\sigma }_1=1\mathrm{and}{\mathrm{\sigma }}_2=1$