Matrix Congruence: Are $B = P^TAP$ and $B = PAP^T$ equivalent?

Question

I know that matrix multiplication is not commutative, but if $B = P^TAP$, then is $B = PAP^T$? In the same way that, if A and B are similar, then $A = P^{-1}BP$ and $A = PBP^{-1}$ are equivalent.

I would appreciate it if someone can please take the time to confirm.

Answer 1

No. If$$P=\begin{pmatrix}1&2\\3&4\end{pmatrix}\text{ and }A=\begin{pmatrix}1&1\\0&1\end{pmatrix},$$then$$P^T.A.P=\begin{pmatrix}13 & 18 \\ 20 & 28\end{pmatrix}\text{, whereas }P.A.P^T=\begin{pmatrix}7 & 15 \\ 17 & 37\end{pmatrix}.$$And it is false with inverses intead of transposes, too.

Answer 2

This is most important for symmetric matrices. What I do is, for the usual case of congruence with a diagonal matrix, is solve $P^T HP = D,$ then $Q = P^{-1} $ solves $Q^T D Q = H.$ Note that the actual eigenvalues of $H$ are irrational, roots of $x^3 - 12x^2 + x + 1.$ Also, $P$ is NOT orthogonal. The diagonal entries of $D$ are subject to Sylvester's Law of Inertia .

$$ H = \left( \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 7 \\ \end{array} \right) $$

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 3 & 0 & 1 \\ - \frac{ 1 }{ 2 } & 1 & - \frac{ 1 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 7 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 3 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & \frac{ 1 }{ 2 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 2 & \frac{ 1 }{ 2 } & 1 \\ 3 & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & \frac{ 1 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & 3 \\ 0 & \frac{ 1 }{ 2 } & 1 \\ 0 & 1 & 0 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 7 \\ \end{array} \right) $$