Adjoints and scalar product. Let $V$ be a vector space over $\mathbb{R}$, $\dim(V)=n<\infty$, $\psi$ a scalar product on $V$ which is not degenerate. Let $f \in \text{End}(V)$, $W \subset V $ such that $f(W) \subset W$. Denote with $f^*$ the adjoint of $f$.
After proving that $W^{\perp}$ is $f^*$-invariant, which means $f^*(W^{\perp})\subset W^{\perp}$ I'm asked to prove that::

The characteristic polynomial of $f^*|_{W^{\perp}}$ divides the characteristic polynomial of $f$.

I really don't know how to proceed and I'd like to receive a few hints that can guide me to the solution. Thanks in advance for your precious help.
Edit: What I've tried.
We know that $A^*=\mathcal{M}_{\mathcal{B}}(f^*) \sim A^t = (\mathcal{M}_{\mathcal{B}}(f))^t \sim A =\mathcal{M}_{\mathcal{B}}(f) $.
So, as the characteristic polynomial of $f^*$ is equal to the characteristic polynomial of $f$ I think that the thesis comes straightforward.
If what I said is true, let's add another hypothesis. Let's say that $\psi|_{W}$ is non degenerate. We want to conclude that the characteristic polynomial of $f$ is the product of the characteristic polynomial of $f|_W$ and the characteristic polynomial of $f^*|_{W^{\perp}}$.
What I've tried to say is: If $\psi |_W$ is non degenerate, we have $V= W \oplus W^{\perp}$. But I don't really know how to proceed.
 A: Consider the direct sum decomposition $V = W \oplus W^{\perp}$ and choose some basis $\beta = (e_1,\dots,e_n)$ such that $(e_1,\dots,e_k)$ is a basis for $W$ and $(e_{k+1},\dots,e_n)$ is a basis for $W^{\perp}$. Since $f(W) \subseteq W$, the matrix representing $f$ with respect to $\beta$ will be block upper triangular
$$ \mathcal{M}_{\beta}(f) = \begin{pmatrix} A & B \\ 0 & C \end{pmatrix} $$
where $A \in M_k(\mathbb{R})$ is the matrix representing $f|_{W}$ with respect to the basis $\beta' = (e_1,\dots,e_k)$ of $W$.  Hence,
$$ \chi_f(\lambda) = \det(A - \lambda I_k) \det(C - \lambda I_{n-k}) = \chi_{f|_{W}}(\lambda) \det(C - \lambda I_{n-k}). $$
Let $G$ be the matrix representing $\psi$ with respect to $\beta$ (so $G_{ij} = \psi(e_i,e_j)$). Because of the direct sum decomposition, the matrix $G$ is block diagonal so write
$$ G = \begin{pmatrix} G_1 & 0 \\ 0 & G_2 \end{pmatrix}. $$
Then
$$ \mathcal{M}_{\beta}(f^{*}) = G^{-1} \begin{pmatrix} A & B \\ 0 & C \end{pmatrix}^t G = \begin{pmatrix} G_1^{-1} & 0 \\ 0 & G_2^{-1} \end{pmatrix} \begin{pmatrix} A^t & 0 \\ B^t & C^t \end{pmatrix} \begin{pmatrix} G_1 & 0 \\ 0 & G_2 \end{pmatrix} =
\begin{pmatrix} G_1^{-1} & 0 \\ 0 & G_2^{-1} \end{pmatrix} 
\begin{pmatrix} A^t G_1 & 0 \\ B^t G_1 & C^t G_2 \end{pmatrix} = \begin{pmatrix} G_1^{-1} A G_1 & 0 \\ G_2^{-1} B^t G_1 & G_2^{-1} C^t G_2\end{pmatrix} $$
and we immediately see that $W^{\perp}$ is $f^{*}$-invariant and $f|_{W^{\perp}}$ is represented with respect to the basis $\beta'' =
 (e_{k+1},\dots,e_n)$ of $W^{\perp}$ by $G_2^{-1} C^t G_2$ so
$$\chi_{f^{*}|_{W^{\perp}}}(\lambda) = \det(G_2^{-1} C^t G_2 - \lambda I_{n-k}) = \det(G_2^{-1} (C - \lambda I_{n-k})^t G_2) = \det(C - \lambda I_{n-k}). $$
Hence, we get $\chi_f(\lambda) = \chi_{f|_{W}}(\lambda) \chi_{f^{*}|_{W^{\perp}}}(\lambda)$.
