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$A$ and $B$ are similar matrices, if $B=PAP^{-1}$ holds for a square, non-singular matrix $P$. Now am wondering if $S^{-1}T$ and $S^{-1/2}TS^{-1/2}$ are similar matrices? Am looking for a proof for it where $S$ is a diagonal matrix. Also- does this similarity hold if $S$ was square but not-diagonal?

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They are similar when $S$ is invertible and diagonal, as long as you fix a choice of $S^{\frac{1}{2}}$. Pre multiply the first by $S^{\frac{1}{2}}$ and postmultiply it by $S^{\frac{-1}{2}}.$ – Geoff Robinson Sep 16 '12 at 7:46
Can you please enter this short multiplication step in the answer section?The reasoning for +1/2 and -1/2? I believe because they cancel each other? Also- does this similarity hold for square but not-diagonal matrices? – user23600 Sep 16 '12 at 7:50
Assuming $S^{\frac{1}{2}}$ exists and is uniquely specified, it will be invertible if $S$ is invertible, and its inverse will be (almost by definition) $S^{\frac{-1}{2}}.$ – Geoff Robinson Sep 16 '12 at 8:30
up vote 0 down vote accepted

$S^{\frac{1}{2}}(S^{-1}T)S^{\frac{-1}{2}} = S^{\frac{-1}{2}}TS^{\frac{-1}{2}}$ as long as $S^{\frac{1}{2}}$ is uniquely specified (assuming it exists- if it didn't, the question would not be meaningful anyway). This does not require $S$ to be diagonal.

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That is: If $R$ is any matrix with the property that $R^2=S$, then $R$ is invertible with invers $RS^{-1}$ because $R\cdot RS^{-1}=SS^{-1}=1$. Then $R(S^{-1}T)R^{-1}=R^{-1}TR^{-1}$. – Hagen von Eitzen Sep 16 '12 at 12:57

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