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MathsGee Android Q&A

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Let \(A\) be a real matrix, not necessarily square.

a) Show that both \(A^{*} A\) and \(A A^{*}\) are self-adjoint.
b) Show that both \(A^{*} A\) and \(A A^{*}\) are positive semi-definite.
c) Show that \(\operatorname{ker} A=\operatorname{ker} A^{*} A\). [HINT: Show separately that \(\operatorname{ker} A \subset \operatorname{ker} A^{*} A\) and ker \(A \supset \operatorname{ker} A^{*} A\). The identity \(\left\langle\vec{x}, A^{*} A \vec{x}\right\rangle=\langle A \vec{x}, A \vec{x}\rangle\) is useful.]
d) If \(A\) is one-to-one, show that \(A^{*} A\) is invertible
e) If \(A\) is onto, show that \(A A^{*}\) is invertible.
f) Show that the non-zero eigenvalues of \(A^{*} A\) and \(A A^{*}\) agree. Generalize.
g) Show that image \(\left(A A^{*}\right)=\left(\operatorname{ker} A A^{*}\right)^{\perp}=\left(\operatorname{ker} A^{*}\right)^{\perp}=\operatorname{image} A\).
in Mathematics by Platinum (147,718 points) | 99 views

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MathsGee Android Q&A

MathsGee Android Q&A