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Let $A$ be an $n\times n$ invertible matrix. I think this is true because I have tried a few different real and complex matrices and they satisfy this.

The trouble I'm having is showing it is true. I started by left multiplying A* to get $A^{*}A+A^{*}(A^{-1})^{*}$

I thought it would be helpful since $A^{*}A$ is invertible, but I'm still stuck.

Perhaps there is a counterexample that I am not seeing.

Daniel P
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E. Roeshink
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1 Answers1

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Note that $(A^{-1})^*=(A^*)^{-1}$. So after you multiply by $A^*$ on the left, you have $$ A^*A+I. $$ This is always invertible, since $A^*A$ is positive semidefinite and so $A^*A+I$ has all its (real) eigenvalues $1$ or greater. So $$ A^*(A+(A^{-1})^*) $$ is invertible. The only way for a product of invertible matrices to be invertible is if both are invertible.

Martin Argerami
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  • Beautiful, thank you. Although, I do believe $A^{*}A$ is positive definite since A is invertible, not that it makes a difference here. – E. Roeshink Apr 19 '19 at 16:55
  • Yes, of course. I come from an operator algebras background, so in my head it's just "positive". I need to "translate" when it comes to talk about matrices. – Martin Argerami Apr 19 '19 at 19:53