# Using Jordan Normal Form to determine when characteristic and minimal polynomials are identical

Say I want to immediately write down a matrix with an identical minimal and characteristic polynomial. Say,

$$(t-1)^{3}(t-2).$$

My first instinct is to write down Jordan Blocks in a block diagonal matrix with blocks $$J_{3}(1), \text{ }J_{1}(2)$$ to give a matrix

$$\begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix}.$$

Clearly this gives the required characteristic polynomial specified above. However, since we know that the minimal polynomial divides the characteristic polynomial, and that distinct roots must show up in the minimal polynomial, using the ease of Jordan Normal Form when multiplying matrices, we can quickly check each case $(t-1)(t-2)$, $(t-1)^{2}(t-2)$, $(t-1)^{3}(t-2)$ and show that the last one is the only such zero matrix required.

However, is it sufficient to argue from the get go, that because we've included only one Jordan Block of each eigenvalue of maximum size, we can immediately say that the characteristic and minimal polynomial are the same (due to the minimal polynomial of each individual block being the same as each of the factors in the characteristic polynomial)? In other words, if the minimal and characteristic polynomial are the same and is

$$\prod_{i=1}^{k} (t- \lambda_{i})^{n_{i}},$$

then a matrix which immediately satisfies this is

$$\begin{bmatrix} J_{n_{1}}(\lambda_{1}) & 0 & \dots & 0 \\ 0 & \ddots & & \\ 0 & 0 & 0 & J_{n_{k}}(\lambda_{k}) \end{bmatrix}.$$

Does this seem reasonable? Thanks in advance.

• you are on right track Mar 20, 2018 at 23:29
• No only right track, it is all perfectly correct. Mar 20, 2018 at 23:30
• @ArnaudMortier I put some equivalent conditions at math.stackexchange.com/questions/92480/… Mar 21, 2018 at 0:33
• @WillJagy Really nice! Mar 21, 2018 at 0:35
• Afiq, the quickest way to cause minimal and characteristic polynomials to coincide is to take a companion matrix; en.wikipedia.org/wiki/Companion_matrix Mar 21, 2018 at 0:54

It is perfectly correct, and the reason is that the exponent of the factor $(X-\lambda)$ in the minimal polynomial is exactly the size of the largest Jordan block associated with $\lambda$.