I am interested in finding a proof of the following property that does not make reference to bases, and ideally doesn't use facts about determinants that depend on the block structure of a matrix.

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> Let $T \in L(V,V)$ be a linear operator on a finite-dimensional space $V$. Suppose $W \preccurlyeq V$ is a $T$-invariant subspace, that is, $T(W) \subset W$. Consider the restriction $T\_W \in L(W,W)$ of $T$ to $W$. Then the characteristic polynomial of $T\_W$ divides the characteristic polynomial of $T$.
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Let $p,p\_W$ be the characteristic polynomials and $m,m\_W$ be the minimal polynomials. It is easy to show "algebraically" that $m\_W \mid m$ since $m$ annihilates $T\_W$, so must be a multiple of the monic generator $m\_W$. However, the only proofs I have seen that $p\_W \mid p$ make use of basis expansions:

* Let $\mathcal{B}=\{ v\_1,\dots,v\_n \}$ be a basis for $V$ such that $\mathcal{B}'=\{ v\_1, \dots, v\_r \}$ form a basis for $W$.
* The matrix of $T$ with respect to $\mathcal{B}$ has the following block form, where $A \in F^{r \times r}$ is the matrix of $T\_W$ with respect to $\mathcal{B'}$, $$[T]\_{\mathcal{B}} = \begin{bmatrix} A & B \\ & C \end{bmatrix} \implies xI - [T]\_{\mathcal{B}} = \begin{bmatrix} xI - A & B \\ & xI-C \end{bmatrix}$$
* Then $p = \det(xI - [T]\_\mathcal{B}) = \det(xI-A)\det(xI-C)$ is a multiple of $p\_W = \det(xI-A)$.

The use of basis expansions and block matrices leaves something to be desired. **Is there a "matrix-free" way to prove this?** Assume we know about Cayley-Hamilton, if it helps.