How to prove the equation below:
$(A^TA + \mu I)^{-1}A^T$ = $A^T(AA^T + \mu I)^{-1}$
no ideas at all, i major in signal processing and it needs some math techniques so i am not familiar with this part of knowledge, really need help please.
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1Have you tried expanding the equation out into two terms? For example, $(A+B)C = AC + BC$. – Matt Groff Feb 14 '17 at 14:49
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Needn't he use the inverse law first – Displayname Feb 14 '17 at 15:13
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At least we know that $(A^TA + \mu I)^{-1}$ is symmetric. – Felix Crazzolara Feb 14 '17 at 15:15
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It is very simple! Left multiply by $(A^TA + \mu I)$, then right multiply by $(AA^T + \mu I)$, you end up with $A^T(AA^T + \mu I)=(AA^T + \mu I)A^T$ which is evidently true by development of LHS and RHS. – Jean Marie Feb 14 '17 at 21:08