On a recent homework assignment I have a question that asks to prove or disprove the following statement.
If $A,B \in \mathbb{M}_n(\mathbb{K})$ are similar, then they are congruent.
A is congruent to B if there exists an invertible matrix P such that:
$A = P^{T} B P$
I suspect that this is false, so I was seeking a counter example but have had a lot of trouble coming up with one. I managed to put this one together:
$A = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 &1 \\ 0 & 1 &1 \end{bmatrix}$ and $B = \begin{bmatrix} 1+\sqrt{2} & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1-\sqrt{2} \end{bmatrix}$
I say that, since these two matrices have the same eigenvalues, they are similar. But since $A$ is symmetric and $B$ is not they cannot be congruent.
Does my logic make any sense or should I consider a different approach?
You counterexample works.
A remark though:
$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ have the same eigenvalues but they are not similar to each other.
Your example works because the eigenvalues are distinct and they share the same set of eigenvalues.