Is this matrix product diagonalizable?

Question

Let $\boldsymbol{A} \in \mathbb{R}^{n \times m}$ and $\boldsymbol{B} \in \mathbb{R}^{n \times n}$ with $m < n$. If $\boldsymbol{A}$ has full column rank and $\boldsymbol{B}$ is diagonalizable, will $\boldsymbol{A}'\boldsymbol{B}\boldsymbol{A}$ be diagonalizable?

Answer 1

No, diagonalisability is not preserved by matrix congruence. E.g. when $m=n=2$, $$ \underbrace{\pmatrix{1&1\\ 0&-1}}_{A^T}\underbrace{\pmatrix{3&-1\\ -2&1}}_B\underbrace{\pmatrix{1&0\\ 1&-1}}_A=\pmatrix{1&0\\ 1&1}. $$ $B$ is diagonalisable over $\mathbb R$ because its characteristic polynomial $x^2-4x+1$ has two distinct real roots. This counterexample extends to other dimensions $2\le m<n$ by considering $$ \pmatrix{A^T&0&0\\ 0&I_{m-2}&0}\pmatrix{B\\ &0\\ &&0}\pmatrix{A&0\\ 0&I_{m-2}\\ 0&0}=\pmatrix{A^TBA&0\\ 0&0}. $$