$A$ 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?
Similar matrix proof
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linear-algebra
matrices
eigenvalues-eigenvectors
numerical-linear-algebra
spectral-theory
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0Assuming $S^{\frac{1}{2}}$ exists and is uniquely specified, it will be invertible if $S$ is invertible, and its inverse will be (almost by definition) $S^{\frac{-1}{2}}.$ – 2012-09-16
1 Answers
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$S^{\frac{1}{2}}(S^{-1}T)S^{\frac{-1}{2}} = S^{\frac{-1}{2}}TS^{\frac{-1}{2}}$ 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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1That 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 $R(S^{-1}T)R^{-1}=R^{-1}TR^{-1}$. – 2012-09-16