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Suppose that $A$ and $B$ are real symmetric matrices and $A$ is invertible. Show that the matrix $A^{-1}B$ is diagonizable(similar to a diagonal matrix) if and only if there exists an invertible $P$ such that $P^{T}AP$ and $P^{T}BP$ are both diagonal matrices.

The 'if' part is straightforward; however the only if part seems quite hard. Can anyone help?

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    $\begingroup$ I suspect that the following reduction is useful: Suppose that $A^{-1}B$ is diagonalizable. Select matrix $P_1$ such that $P_1^TAP_1$ is diagonal with $1$s, followed by $-1$s. We now have $$ (P_1^TAP_1)^{-1}(P_1^TBP_1) = P_1^{-1}(A^{-1}B)P_1. $$ Let $A_1 = P_1^TAP_1,B_1 = P_1^TBP_1, C_1 = P_1^{-1}(A^{-1}B)P_1$. We now have $A_1^{-1}B_1 = C_1$, so $B_1 = A_1C_1$ where $A_1$ has the form described, $B_1$ is symmetric, and $C_1 = A_1^{-1}B_1$ is diagonalizable. $\endgroup$ – Omnomnomnom Dec 29 '19 at 17:55
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    $\begingroup$ At the very least, this leads to a clear proof in the case where $A$ is positive definite since we would then have $A_1 = I$. $\endgroup$ – Omnomnomnom Dec 29 '19 at 17:56
  • $\begingroup$ Note that whenever $A$ has a "square root" $A^{1/2}$, $A^{-1}B$ must be diagonalizable. In particular, $A^{-1}B$ is similar to $A^{-1/2}BA^{-1/2}$, which is symmetric. With that being said: if we allow for complex diagonal matrices, then $A^{-1}B$ will be diagonalizable for any symmetric matrices $A,B$. $\endgroup$ – Omnomnomnom Dec 29 '19 at 18:13
  • $\begingroup$ In this context, it seems that "diagonalizable" specifically means similar to a diagonal matrix with real entries. Can you verify whether this is the intended meaning? $\endgroup$ – Omnomnomnom Dec 29 '19 at 18:14
  • $\begingroup$ @Omnomnomnom Yes, this is the intended meaning of 'diagonal'. $\endgroup$ – j200932 Dec 30 '19 at 2:22
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This is equivalent to part (a) of Theorem 4.5.17 of Horn and Johnson's Matrix Analysis, second edition.

Theorem: Suppose that $A$ and $B$ are Hermitian and $A$ is nonsingular. Let $C = A^{−1} B$. There is a nonsingular $S \in M_n$ and real diagonal matrices $\lambda,M$ such that $A = S\Lambda S^*$ and $B = SMS^*$ if and only if $C$ is diagonalizable and has real eigenvalues.

Your statement amounts to the case where $A$ and $B$ are also real matrices. The proof for the "only-if" direction is as follows:

only-if proof

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  • $\begingroup$ If I find the time (and patience) to do so, I'll try to rewrite this proof to make it a bit more accessible. $\endgroup$ – Omnomnomnom Dec 29 '19 at 18:43
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Hint: Notice that $(P^T A P)^{-1} \cdot P^T B P = P^{-1} A^{-1} B P$ is the product of two diagonal matrices.

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    $\begingroup$ This can sove the 'if' part. What about the 'only if' part? $\endgroup$ – j200932 Dec 29 '19 at 12:43
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Thoughts so far on the "only if" direction:

Suppose that $A^{-1}B$ is diagonalizable. Select matrix $P_1$ such that $P_1^TAP_1$ is diagonal with $1$s, followed by $-1$s. We now have $$ (P_1^TAP_1)^{-1}(P_1^TBP_1) = P_1^{-1}(A^{-1}B)P_1. $$ Let $A_1 = P_1^TAP_1,B_1 = P_1^TBP_1, C_1 = P_1^{-1}(A^{-1}B)P_1$. We now have $A_1^{-1}B_1 = C_1$, so $B_1 = A_1C_1$ where $A_1$ has the form described, $B_1$ is symmetric, and $C_1 = A_1^{-1}B_1$ is diagonalizable.

I suspect that the trick from here is to select an orthogonal $U$ such that $U^TB_1U$ is diagonal and $UA_1 = A_1 U$. Note, however, that $UA_1 = A_1 U$ if and only if $U$ is block-diagonal with the same shape as $A_1 = \operatorname{diag}(I_p,-I_q)$. However, if $B_1$ is diagonalizable with such a $U$, then $B_1$ must also be block-diagonal of the same shape.

Long story short: in order to proceed with the proof in the way that I suspect it should go, we would need to show that if $A_1 = \operatorname{diag}(I_p,-I_q)$, $B_1$ is symmetric, and $A_1^{-1}B_1$ is diagonalizable, then $B_1$ must be block-diagonal (with block sizes $p$ and $q$). However, I don't see a clear reason that this should hold.

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