Minimal polynomial of a 6x6 matrix?

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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?

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3 Answers

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I know this question is old but the answers already posted are really misleading.

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:$$(x-1)(x-3)^2.$$

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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.

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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.

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