Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There are a series of exercises in the book A Polynomial Approach to Linear Algebra which I require some assistance on.

  1. Let $A:V\rightarrow V$ and $B:W\rightarrow W$ be linear operators on finite dimensional vector spaces with minimal polynomials $\mu_A$ and $\mu_B$ respectively. Show that if there exists a surjective linear mapping $Z: V\rightarrow W$ such that $ZA = BZ$ then $\mu_B\mid\mu_A$.

I was able to solve the above problem by noting $ZA^k = B^kZ$ so that $0=Z\mu_A(A)=\mu_A(B)Z=0$. Since $Z$ is surjective, this suggests that $\mu_A(B)=0$ so that $\mu_B\mid\mu_A$. The problem then asks for what conditions does the converse hold, and for that I have no idea how to even approach the problem.

The latter part of the problem asks me to prove the statement for the characteristic polynomial instead of minimal polynomials. For that, I can show $\chi_A(B)=0$ where $\chi_A$ is the characteristic polynomial, but I'm not sure how to conclude that $\chi_B\mid\chi_A$.

Thanks for any assistance.

share|cite|improve this question
up vote 0 down vote accepted

If we replace $W$ with a complement $U$ to $\ker Z$ in $V$, i.e. a subspace $U \subseteq V$ such that $V = U \oplus \ker Z$, we can suppose that $W$ is a linear subspace of $V$ and $Z \colon V \to W$ is the linear projection onto $W$ along the linear subspace $\ker Z$.

From $ZA = BZ$ we easily deduce that $A(\ker Z) \subseteq \ker Z$. You should be able to prove that the endomorphism $Z \circ A \vert_W \colon W \to W$ is similar to $B$. Now pick a basis $\mathcal{B}$ of $\ker Z$ and a basis $\mathcal B'$ of $W$. It is clear that $\mathcal B \cup \mathcal B'$ is a basis of $V$ and, respect to this basis, $A$ is represented by the matrix $$ \begin{pmatrix} M & | & N \\ \hline 0 & | & P \end{pmatrix}, $$ where $M$ represents $A \vert_{\ker Z}$ respect to $\mathcal B$ and $P$ represents $Z \circ A \vert_W$ respect to $\mathcal B'$.

From this it is clear that $\chi_A(t) = \det(M - t \cdot \mathrm{Id}) \cdot \det(P - t \cdot \mathrm{Id}) = \det(M - t \cdot \mathrm{Id}) \cdot \chi_B(t)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.