Prove $PCP^{-1}B^{-1}$ is always symmetric

Question

While working on programming, I noticed this very interesting property, and I've verified it with the code, while I didn't find a way to prove it.

The question is following:

Let $A$ and $B$ be symmetric and invertible matrices. The product $AB$ can be decomposed using Eigenvalue Decomposition (EVD) as $AB = PDP^{-1}$, where $D$ is a diagonal matrix, and $P$ is an invertible matrix.

Given another arbitrary diagonal matrix $C$, prove that the matrix $PCP^{-1}B^{-1}$ is symmetric.

The code to verify:

import numpy as np

# Set the size of the matrices
N = 5

# Generate random symmetric matrix
A = np.random.rand(N, N)
A = A + A.T
B = np.random.rand(N, N)
B = B @ B.T

# EVD of AB
d, P = np.linalg.eig(A @ B)

# Arbitrary diagonal matrix C
C = np.diag(np.random.rand(N))

# Compute the matrix in question
result = P @ C @ np.linalg.inv(P) @ np.linalg.inv(B)

# Check if the result is symmetric
print(np.allclose(result, result.T))

Answer 1

Let us consider first a very particular case. We take $A=B=I$, the identity matrix, and it is enough for our purposes of giving a counterexample to use only $2\times 2$ matrices. We further take $C,P$ as below and compute $PCP^{-1}$: $$ C =\begin{bmatrix}3\\& 4 \end{bmatrix} \ ,\qquad P =\begin{bmatrix}1 & 1\\& 1 \end{bmatrix} \ ,\qquad P^{-1} =\begin{bmatrix}1 & -1\\& 1 \end{bmatrix} \ ,\\ PCP^{-1} = \begin{bmatrix}1 & 1\\& 1 \end{bmatrix} \begin{bmatrix}3 \\& 4 \end{bmatrix} \begin{bmatrix}1 & -1\\& 1 \end{bmatrix} = \begin{bmatrix}3 & 4\\& 4 \end{bmatrix} \begin{bmatrix}1 & -1\\& 1 \end{bmatrix} = \begin{bmatrix}3 & 1\\& 4 \end{bmatrix} \ . $$ This is not a symmetric matrix.

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But when does it work? Why does the claim work randomly so good?

The reason is as follows. It works when the diagonal matrix $C$ has equal diagonal entries on two positions when on the same two positions we have equal entries in $D$. For instance, if $D$ has all diagonal entries different, then the condition is automatically satisfied.

So consider $D$ with diagonal entries $d_1,d_2,\dots,d_n$. Let $C$ have also as a matter of notations the entries $c_1,c_2,\dots,c_n$. In this case, there always exists a polynomial $f$ such that $f(d_k)=c_k$ for all indices $k$ from one to $n$. Then by matrix functional calculus: $$ f(AB) =f(PDP^{-1}) =Pf(D)P^{-1} =PCP^{-1} \ . $$ So we can restate, and want to show that for any polynomial $f$ the matrix $f(AB)B^{-1}$ is symmetric. Such a polynomial is a sum of simple monomials (times coefficients). So it is enough to show this for $f$ in the list $1,x,x^2,x^3,\dots$. (The sum of two symmetric matrices is also symmetric, the product of a scalar and a symmetric matrix is also symmetric.)

Well, for the above values we get the matrices:

• $B^{-1}$ for $f=1$,
• $(AB)B^{-1}=A$ for $f=x$,
• $(AB)^2B^{-1}=ABAB\;B^{-1}=ABA$ for $f=x^2$,
• $(AB)^3B^{-1}=ABABAB\;B^{-1}=ABABA$ for $f=x^3$,

and so on, in general we obtain (for the next cases, the first two are somehow special) the matrix $ABA\dots ABA$, and it is easy to check it is symmetric. (A non-commutative palindrome is applied on symmetric matrices.)