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I know that if $A$ and $B$ either A and B are positive definite matrices then the product matrix $AB$ is similar to $A^{\frac{1}{2}}BA^{\frac{1}{2}}$ and those two matrices have same eigenvalues. This result is also true when $A$ is symmetric positive definite and $B$ is symmetric positive semi-definite matrix.

I am trying to prove that eigenvalues of product matrix $AB$ are same as eigenvalues of $A^{\frac{1}{2}}BA^{\frac{1}{2}}$ when both matrices are symmetric and positive semidefinite. Since matrices are positive semi-definite we cannot use the result about the similarity of matrices.

I tried to prove the result as follows. Suppose $\lambda$ is an eigenvalue of AB we need to show that $\lambda$ is also an eigenvalue of $A^{\frac{1}{2}}BA^{\frac{1}{2}}$. So we have $ABx=A^{\frac{1}{2}}A^{\frac{1}{2}}B x=\lambda x$. I have a hard time convincing myself that $A^{\frac{1}{2}}A^{\frac{1}{2}}B $ and $A^{\frac{1}{2}}BA^{\frac{1}{2}}$ have same eigenvalues. Any references/hints would be helpful.

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  • $\begingroup$ What do you mean by $A^{1/2}$ for a matrix? Is this some sort of matrix square root, or maybe the transpose? $\endgroup$ – Joshua Ruiter Apr 29 '18 at 19:04
  • $\begingroup$ square root of a matrix. $\endgroup$ – user24318 Apr 29 '18 at 20:12
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For any square matrices $X$, $Y$, the matrices $XY$ and $YX$ have the same eigenvalues ( in fact, the same characteristic polynomial). So you can apply you method even if $A$ is not invertible.

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