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.

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 $T$.

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.

  • $\begingroup$ I wonder if a matrix-free proof would generalize to infinite-dimensional vector spaces..? $\endgroup$ Jul 3 '20 at 15:39

The characteristic polynomial does not change if we extend the scalars. So we may assume that the basic field is algebraically closed.

Fact: the exponent of $(x-\lambda)$ in $P_A(x)$ equals the dimension of the subspace $$V_{(\lambda)}\colon =\{v \in V \ | \ (A-\lambda I)^N v= 0 \text{ for some } N \}$$ (the generalized $\lambda$ eigenspace).

Now, if $W\subset V$ is $A$-invariant then clearly $$W_{(\lambda)}\subset V_{(\lambda)}$$ That's enough to prove divisibility.

In fact, if $0\to W \to V \to U\to 0$ is exact sequence of spaces with operator $A$, then $0\to W_{(\lambda)} \to V_{(\lambda)} \to U_{(\lambda)}\to 0$ is exact for all $\lambda$, so we get the product equality for characteristic polynomials in an extension.


Fix $x$. Define $U=xI-T\in L(V,V)$. Since $W$ is a $T$-invariant subspace, we can define $U_W=xI-T_W\in L(W,W)$.

Suppose $x$ is a root of the characteristic polynomial of $T_W$.

Then $\det(xI-T_W)=0$, that is, $\det(U_W)=0$. This means that $U_W$ sends some nonzero subspace of $W$ to $0$. Since $U_W$ is the restriction of $U$ to $W$, we have that $U$ sends some nonzero subspace of $V$ to $0$. Therefore, $\det(U)=0$.

Thus, $\det(xI-T)=0$, and $x$ is a root of the characteristic polynomial of $T$.

Pass to $\Bbb C$ (or the algebraic closure of whatever field we're working in), if we weren't already in an algebraically closed field. A polynomial divides another polynomial iff all the roots of the first are roots of the second. Since every root of $p_W$ is a root of $p$, $~p_W$ divides $p$.

  • 1
    $\begingroup$ Oh, this doesn't quite work if there are repeated roots… $\endgroup$ May 27 '18 at 4:40

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