A question of characteristic polynomials

Question

Let $\pi:\mathfrak g\to \text{End}V$ is a Lie algebra homomorphism and $V,\mathfrak g$ are finite dimensional. Let $U$ be a subspace of $V$ such that $\pi(\mathfrak g)U\subseteq U.$ Define the quotient representation $\tilde{\pi}:\mathfrak g\to \text{End}V/U.$ Then is it true that the characteristic polynomial of $\pi(X)$ is product of characteristic polynomials $\pi(X)|_U$ and $\tilde{\pi}(X)$?

Answer 1

It is true. In general, for linear transformation $ f $ on a finite-dimensional vector space $ V $ and $ f $-stable subspace $ U \subseteq V $, the characteristic polynomial of $ f $ is the product of those of $ f|_U $ and $ \tilde{f} $, where $ \tilde{f} $ is the linear transformation on $ V/U $ induced by $ f $.

Proof. Take a basis $ (v_1, \dots, v_k) $ of $ U $ and extend it to a basis $ (v_1, \dots, v_n) $ of $ V $. Then the matrix expression of $ f $ with respect to the basis $ (v_1, \dots, v_n) $ has the form $$ \begin{pmatrix} A & * \\ 0 & B \end{pmatrix}, $$ where $ A $ is the matrix expression of $ f|_U $ with respect to the basis $ (v_1, \dots, v_k) $ and $ B $ is the matrix expression of $ \tilde{f} $ with respect to the basis $ (v_{k + 1} + U, \dots, v_n + U) $. Hence the characteristic polynomial of $ f $ is \begin{align*} \det\left(TI_n - \begin{pmatrix} A & * \\ 0 & B \end{pmatrix}\right) &= \det \begin{pmatrix} TI_k - A & * \\ 0 & TI_{n - k} - B \end{pmatrix} \\ &= \det(TI_k - A) \det(TI_{n - k} - B), \end{align*} which is the product of the characteristic polynomials of $ f|_U $ and $ \tilde{f} $.

Answer 2

If you have a $m+n$ by $m+n$ block-triangular matrix $$ A=\begin{pmatrix} B & C \\ 0 & D \end{pmatrix} $$ where $B$ is $m\times m$, $C$ is $m\times n$ and $D$ is $n\times n$, then ${\rm det}(A)={\rm det}(B)\times{\rm det}(D)$. I think the property you want will follow from this if you pick a basis of $V$ obtained by completing a basis of $U$.