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Suppose $\Lambda$ is a diagonal matrix, and that $AS=S\Lambda$. $$ \begin{align*} AS=S\Lambda\\ A^{-1}AS=A^{-1}S\Lambda\\ S=A^{-1}S\Lambda\\ S^{-1}=(A^{-1}S\Lambda)^{-1}\\ S^{-1}=\Lambda^{-1}S^{-1}(A^{-1})^{-1}\\ S^{-1}=\Lambda^{-1} S^{-1}A\\ \end{align*} $$

So I got $S^{-1}=\Lambda^{-1} S^{-1}A$.

But if I do it the other way round, $$ \begin{align*} AS=S\Lambda\\ ASS^{-1}=S\Lambda S^{-1}\\ A=S\Lambda S^{-1}\\ S^{-1}A=S^{-1}S\Lambda S^{-1}\\ S^{-1}AA^{-1}=\Lambda S^{-1}A^{-1}\\ S^{-1}=\Lambda S^{-1}A^{-1}\\ \end{align*} $$ To my surprise, I got $S^{-1}=\Lambda S^{-1}A^{-1}$ ! The $A$ in this is an inverse, which didn't happen in the one above. Weird.

Should $S^{-1}=\Lambda^{-1} S^{-1}A$ or $S^{-1}=\Lambda S^{-1}A^{-1}$ be the correct equation? Why is this like that?

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2 Answers 2

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Both are. To see this, note that $AS=S\Lambda$ AND that $\tilde AS=S\tilde \Lambda$ for other matrices $\tilde A$ and $\tilde \Lambda$, namely $\tilde A=A^{-1}$ and $\tilde \Lambda=\Lambda^{-1}$. Hence every relation involving $(A,\Lambda)$ holds with $(\tilde A,\tilde\Lambda)$ as well.

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In the first derivation you got $\Lambda^{-1}$ into the mix (note the last line), not $\Lambda^1$ like in the second one.

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Thanks! Corrected the typo. So now it is: Should $S^{-1}=\Lambda^{-1} S^{-1}A$ or $S^{-1}=\Lambda S^{-1}A^{-1}$ be correct. –  xenon Aug 20 '11 at 7:26
    
Yes. The equivalence of the two expressions can be derived from $A^2=S\Lambda^2 S^{-1}$ via left multiplication by $\Lambda^{-1}S^{-1}$ and right multiplication by $A^{-1}$. –  anon Aug 20 '11 at 7:29
    
So I could simply say $S^{-1}=\Lambda^{-1} S^{-1}A=\Lambda S^{-1}A^{-1}$ that they are all the same? –  xenon Aug 20 '11 at 7:34
    
Yes they are the same. –  anon Aug 20 '11 at 8:09

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