# Minimal polynomial of a 6x6 matrix?

for the matrix $$\begin{pmatrix} 4 & 1 & 0 & 0 & 0 & 0 \\ -1 & 2 & 0 & 0 & 0 & 0\\ 0 & 0 & -1 & -2 & -2 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 4 & 4 & 5 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3 \\ \end{pmatrix}$$

I found the characteristic polynomial to be $(t-1)^2(t-3)^4$. How do I go about finding the minimal polynomial?

edit: after looking through my notes, I found $(t-1)^2(t-3)^2$ as the minimal polynomial. Is this right?

The matrix is a block diagonal matrix. The minimal polynomial of the first block is $$(x-3)^2$$. The minimal polynomial of the second block is $$(x-1)(x-3)$$. The minimal polynomial of the third block is $$x-3$$. Therefore, the minimal polynomial of the big matrix is the least common multiple of these three polynomials, that is: $$(x-1)(x-3)^2.$$

The minimal polynomial must be $(t-1)^r(t-3)^s$ where $1\le r\le 2$ and $1\le s\le 4$. You could try all possibilities.

But if you suspect it is actually $(t-1)^2(t-3)^2$, all you need to check is that $(M-I)^2(M-3I)^2=0$, but that $(M-I)(M-3I)^2=0$ and $(M-I)^2(M-3I)$ are both nonzero.

• That seems so tedious :'( I literally just learned about this stuff last week. But thank you! Oct 30, 2017 at 2:16

You know that the minimal polynomial is a factor of $(t-1)^2(t-3)^4$ by Cayley-Hamilton, so you just need to determine which factors of $(t-1)^2(t-3)^4$ are polynomials that vanish when you plug in your matrix $M$, ie, you need to find which factors of $p(t)$ satisfy $p(M) = 0$. Here the factors are of the form $(t-1)^i(t-3)^j$ with $i \le 2, j \le 4$.

For instance, you would see that $(t-1)^2(t-3)^4$ is forced to be the minimal polynomial were $(M-1)(M-3)^4$ and $(M-1)^2(M-3)^3$ were both found to be nonzero. But if $(M-1)(M-3)^4$ were zero, then the next step would be to check $(M-3)^4$ and $(M-1)(M-3)^3$, and you would proceed similarly until finished.

Alternatively, if you computed the Jordan normal form of your matrix, you would be done. Here the minimal polynomial would be $(t-1)^i(t-3)^j$ with the exponents $i,j$ to be the sizes of the largest Jordan blocks corresponding to eigenvalues $1,3$, respectively.

• Hi, I don't understand this sentence: 'the sizes of the largest Jordan blocks corresponding to eigenvalues 1,3, respectively'. Oct 30, 2017 at 2:23
• @Deanaf if you haven't seen Jordan normal form before I would probably advise ignoring the final paragraph of the answer. In short, an $n \times n$ Jordan block with $\lambda$ the value along the diagonal itself has minimal polynomial $(t-\lambda)^n$, so it follows that you can read off the minimal polynomial for the matrix as a whole from its Jordan normal form in the described fashion. I'm pretty sure that if you don't already have a Jordan normal form computed, it's not an efficient use of time to do so compared to other methods of computing the minimal polynomial. Oct 30, 2017 at 2:37
• I've checked all the combinations, none of them $=0$. My characteristic polynomial must be wrong... Oct 30, 2017 at 2:39
• looks to me that you're right as to the characteristic polynomial. Oct 30, 2017 at 2:47
• My scratch work seems to indicate that $(t-3)^2(t-1)$ is the minimal polynomial, and to prove this one should show that $(M-3I)^2(M-I) =0$ but $(M-3I)^2$ and $(M-3I)(M-I)$ are both nonzero. Oct 30, 2017 at 2:55