Let $A = \begin{pmatrix}a+b & a &0\\a & 2a & a\\0 & a & a+b \end{pmatrix}$ and $B = \begin{pmatrix}a+b & a &0 & 0\\a & 2a & a & 0\\0 & a & 2a & a\\0 & 0 & a & a+b \end{pmatrix}$, where $A$ and $B$ are invertible. Is it possible to find $A^{-1/2}$ and $B^{-1/2}$ such that $A^{-1} = A^{-1/2}A^{-1/2}$ and $B^{-1}=B^{-1/2}B^{-1/2}$?
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It is possible if $A\geq $ and $B\geq 0$ – Student Jan 25 '18 at 12:51
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I mean that $\langle Ax\mid x\rangle \ge 0$ for all $x$ – Student Jan 25 '18 at 12:57
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yes they are positive definite. – Satya Prakash Jan 25 '18 at 12:59
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So $A^{1/2}$ exists and we have $A=A^{1/2}A^{1/2}$ – Student Jan 25 '18 at 13:01
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Yes. But is it possible to find analytical expressions for them? – Satya Prakash Jan 25 '18 at 13:02
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You should try to find a solution. – Student Jan 25 '18 at 13:10