# Simultaneous diagonalization of bilinear form and a linear transformation

Suppose that $V$ is a finite dimensional $\mathbb R$-vector space, and $A\in M_n(\mathbb R)$, given $T:V\rightarrow V$ a linear transformation and $f:V\rightarrow \mathbb R$ a bilinear form, and $\alpha$ a basis where the matrix representation of $T$ and $f$ is $A$. Is true that if there is a basis $\beta$ where $f$ is diagonal, then $T$ is diagonal in this basis too?

I thought that if $P$ is the basis change matrix from $\beta$ to $\alpha$, we have $D=P^{-1}AP$ in this basis, and $D$ is the representation of $f$ and $T$ in $\beta$. Is this right?

Hint: $$A = \begin{bmatrix} 0&1\\ 4&0\end{bmatrix}.$$