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?

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

Yes, both of these are true in general.

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 $$ p(J) = 0 \iff (x - \lambda)^q \mid p(x) $$

| cite | improve this answer | |

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.