Characteristic polynomial of operator is the product of characteristic polynomials of two induced operators (restricted and quotient space operators).

Question

Consider a linear operator $T: V \to V$ on a finite-dimensional vector space $V$. Suppose that $U$ is a $T$-invariant subspace of $V$, and consider the map $T/U: V/U \to V/U$ given by $T/U(v + U) = T(v) + U$ (here $V/U$ is the quotient space). I'm interested in how exactly the characteristic polynomials of the three operators $T/U, T,$ and $T|_U$ are related (here $T|_U$ denotes the operator $T$ restricted to $U$). I can show that the eigenvalues of $T|_U$ and the eigenvalues of $T/U$ are also eigenvalues of $T$, so my conjecture is that $p_T(\lambda) = p_{T|_U}(\lambda) \cdot p_{T/U}(\lambda)$, where in general $p_\ell(\lambda)$ is the characteristic polynomial of the linear operator $\ell$. However, I'm not sure how to go about showing that the multiplicities of each eigenvalue "match up".

Answer 1

Choose a basis of $U$, and complete it into a basis of $V$. On the other hand, the vectors that you used to complete the basis can be used to form a basis of $V/U$.

The matrix of $T$ in the completed basis then looks like $$\left[\matrix{M_{T_{U}} &*\\0&M_{T/U}}\right]$$ where $M_{T/U}$ is with respect to the basis from the remark above, and $*$ stands for irrelevant (though possibly non-zero) entries.

Your conjecture follows immediately.