Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

If A is an invertible $2 x 2$ matrix, what is the relationship between the singular values of A and $A^{- 1}$ ? Justify your answer in terms of the image of the unit circle.

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: To find the singular values of A.

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

${(A^{- 1})}^{t} A^{- 1} = {(A^{t})}^{- 1} A^{- 1} = {({AA}^{t})}^{- 1}$

Since $σ^{2}$ is an eigenvalue of ${({AA}^{t})}^{- 1}$ and since ${({AA}^{t})}^{- 1}$ is an invertible matrix, therefore $σ^{2} \neq 0$ Now we know that, if $λ$ is an eigenvalue of an invertible matrix A, then $\frac{1}{λ}$ is an eigenvalue of $A^{- 1}$ Therefore $\frac{1}{σ^{2}}$ is an eigenvalue of ${({({AA}^{t})}^{- 1})}^{- 1} = AA$

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

$(A^{t} A) v = λv$

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

$({AA}^{t}) (Av) = λ (Av) . . . . . . . . . . . . (l)$

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

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

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

Let A be an invertible $2 \times 2$ matrix. Also, let $σ_{1} σ_{2}$ 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 $σ_{1}$ and $σ_{2}$ respectively. Now consider an ellipse whose semi-major and semi-minor axes have lengths $\frac{1}{σ_{2}}$ and $\frac{1}{σ_{1}}$ 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 $A^{- 1}$ is this ellipse, which means the ellipse whose semi-major and semi-minor axes have lengths $\frac{1}{σ_{2}}$ and $\frac{1}{σ_{1}}$ respectively. Therefore, by using Theorem 8.3.2 we get that, the singular values of $A^{- 1}$ are $\frac{1}{σ_{2}}$ and $\frac{1}{σ_{1}}$

Step 2: Final proof

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

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Linear Algebra With Applications

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Short Answer

If A is an invertible $2 x 2$ matrix, what is the relationship between the singular values of A and $A^{- 1}$ ? Justify your answer in terms of the image of the unit circle.

Step by Step Solution

Step 1: To find the singular values of A.

Step 2: Final proof

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Linear Algebra With Applications

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If A is an invertible 2x2 matrix, what is the relationship between the singular values of A and A-1? Justify your answer in terms of the image of the unit circle.

Step by Step Solution

Step 1: To find the singular values of A.

Step 2: Final proof

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If A is an invertible $2 x 2$ matrix, what is the relationship between the singular values of A and $A^{- 1}$ ? Justify your answer in terms of the image of the unit circle.