Finding the minimal polynomial without using the characteristic polynomial

linear algebramatricesminimal-polynomials

I want to find the minimal polynomial (the monic polynomial of least positive degree that annihilates the matrix) of the following matrix:
$$\begin{pmatrix}
0 & 1 & 1 & 0\\
-1 & 0 & 0 & 1\\
-1 & 0 & 0 & 1\\
0 & -1 & -1 & 0
\end{pmatrix}
$$

I know how to use the characteristic polynomial to find the minimal polynomial, but is there a way to get around that? It's not quite symmetric, so I can't use the trick that applies to symmetric matrices.

Thank you.

Best Answer

The matrix is skew-symmetric and one can write it as

$$M=\begin{bmatrix}r_1\\r_2\\r_2\\-r_1\end{bmatrix}$$

where the rows $r_1$ and $r_2$ are immediate from the definition. Since it is skew-symmetric, then we also have that

$$M=\begin{bmatrix}-r_1^T & -r_2^T&-r_2^T&r_1^T\end{bmatrix}.$$

We have that $r_1^Tr_2=0$ and $r_1^Tr_1=r_2^Tr_2=2$ together with $r_1M=2r_2$ and $r_2M=-2r_1$.

Therefore,

$$M^2=2\begin{bmatrix}r_2\\-r_1\\-r_1\\-r_2\end{bmatrix}\ \mathrm{and}\ M^3=-4\begin{bmatrix}r_1\\r_2\\r_2\\r_1\end{bmatrix}=-4M.$$

Therefore, the minimal polynomial is $p(x)=x^3+4x=x(x^2+4)$.

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