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Expert-verified Found in: Page 412 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # If A is an invertible ${\mathbf{2}}{\mathbf{x}}{\mathbf{2}}$ matrix, what is the relationship between the singular values of A and ${{\mathbf{A}}}^{\mathbf{-}\mathbf{1}}$? Justify your answer in terms of the image of the unit circle.

If $\sigma$ is a singular value of A, then $\frac{1}{\sigma }$ is a singular value of ${A}^{-1}$

See the step by step solution

## Step 1: To find the singular values of A.

Let A be an invertible $2×2$ matrix. Let $\sigma$ be a singular value of ${\mathrm{A}}^{-1}$ Then by definition, ${\sigma }^{2}$ is an eigenvalue ${\left({\mathrm{A}}^{-1}\right)}^{\mathrm{t}}{\mathrm{A}}^{-1}$ of Now we know

${\left({\mathrm{A}}^{-1}\right)}^{\mathrm{t}}{\mathrm{A}}^{-1}={\left({\mathrm{A}}^{\mathrm{t}}\right)}^{-1}{\mathrm{A}}^{-1}={\left({\mathrm{AA}}^{\mathrm{t}}\right)}^{-1}$

Since ${\sigma }^{2}$ is an eigenvalue of ${\left({\mathrm{AA}}^{\mathrm{t}}\right)}^{-1}$ and since ${\left({\mathrm{AA}}^{\mathrm{t}}\right)}^{-1}$ is an invertible matrix, therefore ${\sigma }^{2}\ne 0$ Now we know that, if $\lambda$ is an eigenvalue of an invertible matrix A, then $\frac{1}{\lambda }$ is an eigenvalue of ${A}^{-1}$ Therefore $\frac{1}{{\sigma }^{2}}$ is an eigenvalue of ${\left({\left({\mathrm{AA}}^{\mathrm{t}}\right)}^{-1}\right)}^{-1}=\mathrm{AA}$

Also, we know that ${\mathrm{AA}}^{\mathrm{t}}$ and ${\mathrm{AA}}^{\mathrm{t}}$ have the same nonzero eigenvalues for any matrix A. [Proof: Let $\lambda \ne 0$ is an eigenvalue of ${\mathrm{A}}^{\mathrm{t}}\mathrm{A}$ then there exists a nonzero vector v such that

$\left({\mathrm{A}}^{\mathrm{t}}\mathrm{A}\right)\mathrm{v}=\mathrm{\lambda v}$

Now if we multiply both sides of the above equation by A, then we get

$\left({\mathrm{AA}}^{\mathrm{t}}\right)\left(\mathrm{Av}\right)=\mathrm{\lambda }\left(\mathrm{Av}\right)............\left(\mathrm{l}\right)$

If then equation (l) implies that $\mathrm{\lambda v}=0$. Now as $\lambda \ne 0$ and v is a nonzero vector, therefore $\mathrm{\lambda v}$ cannot be zero. Hence Av is a nonzero vector. Therefore, is an eigenvalue of ${\mathrm{AA}}^{\mathrm{t}}$

Thus $\frac{1}{{\sigma }^{2}}$ is also an eigenvalue of ${\mathrm{A}}^{\mathrm{t}}\mathrm{A}$ This implies that $\sqrt{\frac{1}{{\sigma }^{2}}}=\frac{1}{\sigma }$ is a singular value of A.

Thus we get that if is a singular value of A, then $\frac{1}{\sigma }$ is a singular value of ${A}^{-1}$

Let A be an invertible $2×2$ matrix. Also, let ${\sigma }_{1}{\sigma }_{2}$ be the singular values of A. Now by using Theorem 8.3.2 we get that, the image of the unit circle under A is an ellipse whose semi-major and semi-minor axes have lengths ${\sigma }_{1}$ and ${\sigma }_{2}$ respectively. Now consider an ellipse whose semi-major and semi-minor axes have lengths $\frac{1}{{\mathrm{\sigma }}_{2}}$ and $\frac{1}{{\mathrm{\sigma }}_{1}}$ respectively. Then similarly by using Theorem 8.3.2 we get that, the image of this ellipse under A is the unit circle. Hence by the definition of ${A}^{-1}$ the image of the unit circle under ${\mathrm{A}}^{-1}$ is this ellipse, which means the ellipse whose semi-major and semi-minor axes have lengths $\frac{1}{{\mathrm{\sigma }}_{2}}$ and $\frac{1}{{\mathrm{\sigma }}_{1}}$ respectively. Therefore, by using Theorem 8.3.2 we get that, the singular values of ${\mathrm{A}}^{-1}$ are $\frac{1}{{\mathrm{\sigma }}_{2}}$ and $\frac{1}{{\mathrm{\sigma }}_{1}}$

## Step 2: Final proof

If $\sigma$ is a singular value of A, then $\frac{1}{\sigma }$ is a singular value of ${\mathrm{A}}^{-1}$ ### Want to see more solutions like these? 