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If $A,B$ are symmetric $m$-by-$m$ matrices with $A$ positive definite, I want to show that $A^{1/2}BA^{1/2}$ is a diagonal matrix if it is given that $BAB=B$. Here $A^{1/2}$ is the positive definite square root of $A$.

I have been stuck on this problem for a long time. I have tried the decomposition of $A,B$ into the products of orthogonal matrices and diagonal matrices, but maybe the correct path has not hit me. I have not yet arrived at any notable result in trying to prove this. All my attempts have failed.

I would really appreciate some help.

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    $\begingroup$ A matrix such that its square equals $A$, not the. $\endgroup$ – Git Gud Feb 9 '15 at 16:39
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    $\begingroup$ $A$ is positive definite. $\endgroup$ – Landon Carter Feb 9 '15 at 16:39
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    $\begingroup$ Given $BAB = B$ we have $BA^{1/2}A^{1/2}B = B$ and multiplying by $A^{1/2}$ on both sides we find $(A^{1/2}BA^{1/2})^2 = A^{1/2}BA^{1/2}$. In other words $A^{1/2}BA^{1/2}$ is idempotent. A symmetric idempotent matrix is a projection matrix. $\endgroup$ – Joel Feb 9 '15 at 16:40
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    $\begingroup$ @yedaynara there can be several "square roots" of a matrix. For instance, the 2x2 identity, $I_2$ has $I_2$ as a square root as well as $diag(-1,-1)$, $diag(1,-1)$, and $diag(-1,1)$. (Here diag is a matrix with diagonal entries given and zero entries otherwise.) $\endgroup$ – Joel Feb 9 '15 at 16:47
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    $\begingroup$ Do you think that $A^{1/2}BA^{1/2}$ will be really diagonal? But if you take any idempotent symmetric matrix $B$, then $A=I$ satisfies the given conditions, without making $A^{1/2}BA^{1/2}$ diagonal. $\endgroup$ – Samrat Mukhopadhyay Feb 9 '15 at 17:36
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Given $BAB = B$ we have $BA^{1/2}A^{1/2}B = B$ and multiplying by $A^{1/2}$ on both sides we find $(A^{1/2}BA^{1/2})^2 = A^{1/2}BA^{1/2}$. In other words $A^{1/2}BA^{1/2}$ is idempotent.

A symmetric idempotent matrix is a projection matrix.

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    $\begingroup$ Sorry I also proved that it is idempotent but I do not know definition of projection matrix. Although I understand what a projection matrix means, I do not really understand whether it has to be diagonal. $\endgroup$ – Landon Carter Feb 9 '15 at 16:47
  • $\begingroup$ Projection matrices are not necessarily diagonal. They are diagonalizable. This is meant as a hint. $\endgroup$ – Joel Feb 9 '15 at 16:48
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    $\begingroup$ LOL I do not know about these things. Anyway I guess your answer really answers my question though apparently I do not understand how a non-diagonal matrix can be the result of $A^{\frac{1}{2}}BA^{\frac{1}{2}}$ when I want the result to be diagonal. $\endgroup$ – Landon Carter Feb 9 '15 at 16:52
  • $\begingroup$ So, what would you suggest me to look up? Diagonalization of a matrix? I do not know anything about it. But I do really hope that $A^{\frac{1}{2}}BA^{\frac{1}{2}}$ turns out to be diagonal. $\endgroup$ – Landon Carter Feb 9 '15 at 17:00
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    $\begingroup$ Hello @yedaynara, thanks for the question. In the future please make your title specifically address the question you have. Do not put things such as "Help me", "Question on", "stuck on", etc. in the title. I have fixed this title for you. Regards $\endgroup$ – 6005 Feb 9 '15 at 19:42
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The claim is not true in general. Take $$ B=\pmatrix{1&0\\0&0}.$$ Let $A$ be symmetric positive definite mit $a_{11}=1$. Then $BAB=B$.

If $A$ is not diagonal, then $A^{1/2} = \pmatrix{c_{11}&c_{12}\\c_{12} & c_{22}}$ is not diagonal, and $$ A^{1/2} B A^{1/2} = \pmatrix{c_{11}&c_{12}\\c_{12} & c_{22}} \pmatrix{1&0\\0&0} \pmatrix{c_{11}&c_{12}\\c_{12} & c_{22}} = \pmatrix{c_{11}&0\\c_{12}&0}\pmatrix{c_{11}&c_{12}\\c_{12} & c_{22}} = \pmatrix{c_{11}^2&c_{11}c_{12}\\ c_{11}c_{12}&c_{12}^2} $$ is not diagonal either.


It in addition, $B$ is assumed to be invertible, then the claim follows. Multipyling the equation $BAB=B$ with $B^{-1}$ yields $$ AB=BA =I. $$ Thus, $A=B^{-1}$, and $B$ is positive definite as well.

Denoting by $A^{1/2}$ the (uniquely determined) symmetric and positive definite square root of $A$, it follows, $A^{1/2} = (B^{-1})^{1/2} = (B^{1/2})^{-1}$. Hence $$ A^{1/2} B A^{1/2} = I, $$ which is diagonal.

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  • $\begingroup$ Well done @daw. After posting my answer, I started to believe that there wasn't much else you could really claim other than that the matrix in question was a projection matrix. This demonstrates that the result cannot be improved. $\endgroup$ – Joel Feb 9 '15 at 20:40

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