I came across a problem recently in my linear algebra studies that went something like this:

Let $A$ be a linear transformation on a finite-dimensional space $V$ with characteristic polynomial $(x - 1)^6(x + 6)^2(x - 9)$ and minimal polynomial $m_A(x) = (x - 1)(x + 6)(x -9)$. Let $B$ be a linear transformation which commutes with $A$ and has characteristic polynomial $x^3(x - 2)^6$. Describe all possible minimal polynomials of $B$.

Now, via Cayley-Hamilton, I know that $m_B(x)$ must divide $x^3(x - 2)^6$, so $m_B(x) = x^a(x - 2)^b$, where the possibilities for $a$ are $1, 2, 3$ and for $b$ are $1, 2, 3, 4, 5, 6$, giving us a total of $18$ initial possibilities for the minimal polynomial. Moreover, I know that $A$ and $B$ are simultaneously triangulable, since they commute and operate on a finite-dimensional space $V$. Furthermore, I know $A$ is diagonalizable, since its minimal polynomial has only simple roots; and $A$ is invertible, since none of its eigenvalues is zero.

I would like to straightforwardly limit my choices for $m_B(x)$, but I cannot see a simple approach to eliminating any choices for $a$ and $b$. I tried to see what would happen if $a = b = 1$, but I got nowhere with this. Any ideas as to how I should limit my choices for $a$ and $b$?


So you know that $A$ has eigenspaces of dimensions $6,2,1$. The main constraint is that each of these subspaces is $B$-stable since $A$ and $B$ commute. This means that each of the eigenspaces for $A$ can be separately decomposed into Jordan blocks for $B$.

Now you know that the Jordan blocks for $B$ add up in size to $3$ for $\lambda=0$, and to $6$ for $\lambda=2$. This means for instance that if there is a single Jordan block of size$~3$ for $\lambda=0$, then it must be inside the eigenspace of dimension$~6$ for$~A$, and this only leaves space for Jordan blocks of size at most$~3$ for $\lambda=2$. So if in your terms $a=3$ then $1\leq b\leq3$. On the other hand if $a<3$, then its Jordan block can be lodged inside the $2$-dimensional eigenspace, and one only can say $1\leq b\leq6$ as you already knew.

  • $\begingroup$ I like this approach. It appeals to the basic behaviors of Jordan blocks and simply works with relative sizes and what blocks can "fit." Thank you!! $\endgroup$ – d4rk_1nf1n1ty Apr 16 '15 at 5:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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