# What is the connection between Jordan Canonical Form and minimal polynomial?

I saw a bunch of examples of Jordan Canonical Form and how it is related to the minimal polynomial. I have noticed the following patterns:

Let $A$ be a matrix and $\lambda_1,\dots,\lambda_k$ be the eigenvalues of $A$.

(1). The number of Jordan blocks corresponding to the eigenvalue $\lambda_j$ is the dimension of the eigenspace of $\lambda_j$.

(2). The power of $(x-\lambda_j)$ in the minimal polynomial of $A$ is the size of the largest Jordan block corresponding to $\lambda_j$.

Are (1) and (2) true in general?

• If $B$ is upper triangular and also zero on the diagonal then what is its minimal polynomial ? – reuns Aug 4 '17 at 4:36
• Yes, they are true in general. – zipirovich Aug 4 '17 at 4:37

To see that (1) is true, note that if $$J$$ is in Jordan form, then $$\dim(\ker(J - \lambda I))$$ will simply be the number of $$0$$-columns in $$J - \lambda I$$. Note that these $$0$$-columns only occur at the start of any Jordan block.
To see that (2) is true, note that for any polynomial $$p$$, $$p(J) = p(J_1) \oplus p(J_2) \oplus \cdots \oplus p(J_m)$$ where each $$J_i$$ denotes are Jordan block, and $$\oplus$$ denotes a diagonal direct sum. Note that if $$J$$ is the Jordan block of size $$q$$ associated with $$\lambda$$, then we will have $$p(J) = 0 \iff (x - \lambda)^q \mid p(x)$$