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Exercise 6.7.19

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1.

The nonzero singular values of

A

are the positive square roots of the nonzero eigenvalues of

A^{∗} A

. But the eigenvalues of

A^{∗} A

and that of

A A^{∗}

are the same by Exercise 6.7.9(c). Hence we know that the singular value decomposition of

A

and that of

A^{∗}

are the same.

Also, we have

{(A^{t})}^{∗} A^{t} = \bar{A} {\bar{A}}^{∗} = \bar{A A^{∗}} .

Since $A A^{∗}$ is self-adjoint, its eigenvalues are always real. We get that if $A A^{∗} (x) = λx$ , then we have

{(A^{t})}^{∗} A^{t} (\bar{x}) = \bar{A A^{∗} (x)} = \bar{λ} \bar{x} = λ \bar{x},

here $\bar{A}$ means the matrix consisting of the conjugate of the entries of $A$ . Hence the singular value of $A^{t}$ and that of $A^{∗}$ are all the same.

2.

Let

A = U Σ V^{∗}

be a singular value decomposition of

A

. Then we have

A^{∗} = V Σ^{∗} U^{∗}

. So

{(A^{∗})}^{†} = U {(Σ^{∗})}^{†} V^{∗} = U {(Σ^{†})}^{∗} V^{∗} = {(A^{†})}^{∗} .

3.

Let

A = U Σ V^{∗}

be a singular value decomposition of

A

. Then we have

A^{t} = {(V^{∗})}^{t} Σ^{t} U^{t}

. So

{(A^{t})}^{†} = {(U^{t})}^{∗} {(Σ^{t})}^{†} V^{t} = {(U^{∗})}^{t} {(Σ^{†})}^{t} V^{t} = {(A^{†})}^{t} .

jephianlin

2011-06-27 00:00

Exercise 6.7.19

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