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How do I shown that the matrices $AB$ and $A^{1/2}B A^{1/2}$ have same eigenvalues? Here both $A$ and $B$ are symmetric matrices and $A^{1/2}$ is the square root of matrix $A$.

This book mentions the relation in Remark 4.2

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I think you need $A$ to be nonnegative definite. –  matgaio Apr 25 '12 at 20:14
    
@matgaio Yes you are correct. Both $A$ and $B$ are positive semi-definite. –  sauravrt Apr 25 '12 at 20:15
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2 Answers 2

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Using the fact that $\textrm{spec}(AB)=\textrm{spec}(BA)$, it follows directly from

$$\textrm{spec}(A^{1/2}BA^{1/2})=\textrm{spec}(A^{1/2}A^{1/2}B)=\textrm{spec}(AB)$$

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Do mean eigenvalues by spec() ? –  sauravrt Apr 25 '12 at 20:26
    
Yes, the name of the set of eigenvalues of an operator in finite dimension is Spectrum, then the notation. –  matgaio Apr 25 '12 at 20:30
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There is a classical result which says that if $M_1$ and $M_2$ are two square matrices of the same dimension, then eigenvalues of $M_1M_2$ are the same as the eigenvalues of $M_2M_1$. To show this, first show if when $M_1$ is invertible, and to jump to the general case approximate $M_1$ by invertible matrices.

Indeed, when $M_1$ is invertible, we have $$\det(M_1M_2-xI)=\det M_1(\det M_2-xM_1^{-1})=\det(M_2M_1-xI),$$ and if $M_1$ is not invertible, $M_1^{(n)}:=M_1+n^{-1}I$ is invertible for $n$ large enough so $$ \det(M_1^{(n)}M_2-xI)=\det(M_2M_1^{(n)}-xI),$$ and use the continuity of the determinant.

Now apply this to $M_1=A^{1/2}$ and $M_2=A^{1/2}B$.

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Given that eigenvalues of $M_1M_2$ and $M_2M_1$ are same, the description given by @matgaio is straightforward. Thanks. –  sauravrt Apr 25 '12 at 20:31
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