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I have two square matrices $A=\begin{bmatrix} a_{11}&a_{12}&0&0\\ a_{21}&a_{22}&0&0\\ 0&0&a_{33}&a_{34}\\ 0&0&a_{43}&a_{44} \end{bmatrix}$ and $B=\begin{bmatrix} b_{11}&0&0&0\\ 0&b_{22}&0&0\\ 0&0&b_{33}&0\\ 0&0&0&b_{44} \end{bmatrix},$which are in product form as shown in the following equation \begin{equation} A^{-1}BA \end{equation} where $B$ is a diagonal positive definite matrix. $A$ can be positive, negative or indefinite matrix.

If the main diagonal elements $B$ are equal (i.e., B is scalar positive definite), then obviously $A^{-1}BA=BI,$ which is a positive definite. But I am not sure what will be the result if $B$ is a diagonal with different elements on its main diagonal?

Nihad Ali
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  • There is no reason to believe that $A^{-1}BA$ is symmetric, let alone positive definite. In fact, $A^{-1}BA$ is symmetric if and only if $A^2$ commutes with $B$. If the diagonal entries of $B$ are distinct, then $A^{-1}BA$ is symmetric if and only if $A^2$ is diagonal. – Ben Grossmann Dec 20 '20 at 19:13
  • @BenGrossmann I am wondering why is the result of $A^{-1}BA$ needs to be symmetric? I was asking about positive definiteness of the resultant, and asymmetric matrix can be positive definite. For example \begin{bmatrix} 5&1\ 2& 3 \end{bmatrix}, is asymmetric but positive definite – Nihad Ali Dec 21 '20 at 13:45
  • There are two common definitions for positive definite. One definition requires that the matrix is symmetric, the other does not. Because the matrices you started with are both symmetric, I assumed you were using the first definition. – Ben Grossmann Dec 21 '20 at 13:51
  • With that said, this post seems relevant – Ben Grossmann Dec 21 '20 at 13:54
  • For my case $A$ is asymmetric. I got an idean, help me if am wrong. The product of two matrices (1) PD$\times$PD=PD (2) ND$\times$PD=ND (3) ND$\times$ND=PD where PD and ND represent Positive and Negative definite matrices respectively. if A is PD, then $A^{-1}$ is also PD, and the resultant will be like PDPDPD=PD if A is ND then $A^{-1}$ is also $ND$, and the resultant will be like NDPDND=PD – Nihad Ali Dec 21 '20 at 14:03
  • As the post I linked demonstrates, it is not true that $PD \times PD = PD$. – Ben Grossmann Dec 21 '20 at 14:08

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Counterexample: Take $$ A = \pmatrix{2&1&0&0\\1&2&0&0\\0&0&1&0\\0&0&0&1}, \quad B = \pmatrix{10\\&1\\&&1\\&&&1}. $$ We find that $$ A^{-1}BA = \pmatrix{13&6&0&0\\-6&-2&0&0\\0&0&1&0\\0&0&0&1} $$ has a negative diagonal entry and therefore cannot be positive definite.

Ben Grossmann
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  • I think the signs of the elements on diagonal has decides negative of positive definiteness of a matrix IFF the matrix is Diagonal matrix. In the counter example you provided the resultant is positive definite. I checked on MATLAB – Nihad Ali Dec 21 '20 at 14:16
  • @NihadAli What exactly is your definition of positive definite? Typically, we say that $M$ is positive definite if and only if for every non-zero column-vector $x$, $x^TMx > 0$. However, with $M = A^{-1}BA$ as above, taking x = [0;1;0;0] yields $x^TMx = -2 < 0$ – Ben Grossmann Dec 21 '20 at 14:20
  • @NihadAli Note that for a non-symmetric matrix $M$, it is no longer true that $M$ is positive deifnite if and only if its eigenvalues are positive. – Ben Grossmann Dec 21 '20 at 14:21
  • I check definiteness of a matrix based on eigen values..

    If all eigen values are positive then matrix is positive....

    If all negative then negative..

     – Nihad Ali Dec 21 '20 at 14:39
  • @NihadAli That is not correct. I strongly recommend that you review your notes for the definition of "positive definite". – Ben Grossmann Dec 21 '20 at 14:56
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    Yeah you were right, eigen values procedure is valid for symmetric matrices and may not necessarily valid for non-symmetric matrices. I am really thankful for your precious time. Thanks alot. – Nihad Ali Dec 21 '20 at 15:02
  • @NihadAli Glad to help – Ben Grossmann Dec 21 '20 at 15:03