Minimal polynomial from Jordan canonical form.

Question

Suppose $T:V\to V$ is a linear operator on $V$ which is finite dimensional and suppose the JCF of $T$ is,

$\begin{pmatrix}c & \\1 &c\\ & &c\\ & & &d\\ & & & 1&d\\ & & & &1 &d\end{pmatrix}_{6\times 6}$,then clearly the characteristic polynomial of $T$ is $c_T(X)=(X-c)^3(X-d)^3$ i.e. the power of $(X-c)$ is the number of $c$'s in the diagonal and same way for $(X-d)$.But I do not think it is easy to find the minimal polynomial.Clearly,it is of the form $(X-c)^\alpha (X-d)^\beta$ where $1\leq \alpha,\beta\leq 3$,but we cannot say anything more.But our linear algebra professor told us that to find the minimal polynomial we have to just look at the largest Jordan blocks corresponding to a given eigenvalue,so for this example, the minimal polynomial should be $m_T(X)=(X-c)^2(X-d)^3$ but it does not annihilate $T$,so what is going wrong here?Can someone please clarify?

Answer 1

the minimal polynomial is the product of $(x- \lambda_i)^{M_i} $ where $M_i$ is the size of the largest Jordan block for the eigenvalue $\lambda_i$

Answer 2

The Jordan canonical form describe the action of a linear map dividing it in blocks of invariant subspaces. Clearly your map is divided in the action over three different invariant subspaces, as it have three distinct Jordan blocks, so you can write your transformation $T$ as $T=T_1+T_2+T_3$ where each $T_j$ have one of the Jordan blocks and is zero in all the other entries, so $T_j^n T_k^m=0$ when $j\neq k$ and $n,m\geqslant 1$.

In your case

$$ \begin{bmatrix} c&&&&&\\1&c&&&&\\&&c&&&\\&&&d&&\\&&&1&d&\\&&&&1&d \end{bmatrix}= \begin{bmatrix} c&&&&&\\1&c& \ & \ &\ & \ \\&&&&&\\&&&&&\\&&&&&\\&&&&& \\ \end{bmatrix}+\begin{bmatrix} &&&&&\\&&&&&\\ \ &\ &c&\ &\ & \\ \\ \\ \\ \end{bmatrix}+\begin{bmatrix} &&&&&\\&&&&&\\&&&&&\\&&&d&&\\ \ &\ &\ &1&d&\ \\&&&&1&d \end{bmatrix} $$

From there its easy to see that if $p(T)=p(T_1+T_2+T_3)=0$ is minimal then $p(x)=(x-c)^2 (x-d)^3$, so you did some miscalculation computing $p(T)$ if it doesn't gives zero.

To be more precise: note that $(T-cI)^2$ and $(T-dI)^3$ gives two matrices that act in completely different invariant subspaces (to see it divide $I=I_1+I_2+I_3$ where each $I_j$ have just positive entries in the diagonal where there are non-zero entries in $T_j$) so it product is necessarily zero.