# Have I found the Jordan form correctly?

I am given that the minimal polynomial and characteristic polynomial of a matrix are both $(x-1)^2(x+1)^2$. I have found the Jordan form to be $$\begin{bmatrix}1&1&0&0\\0&1&0&0\\0&0&-1&1\\0&0&0&-1\end{bmatrix}.$$ Is this correct or have I made a mistake somewhere?

• That's perfect. – Arnaud Mortier Jun 26 '18 at 18:17

## 2 Answers

The Jordan form has to be \begin{bmatrix} 1 & a & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & b \\ 0 & 0 & 0 & -1 \end{bmatrix} because of the multiplicities, with $a,b\in\{0,1\}$. If $a=0$, the upper $2\times2$ block satisfies the polynomial $x-1$, so the matrix satisfies $(x-1)(x+1)^2$. Similarly, if $b=0$, the matrix satisfies $(x-1)^2(x+1)$. Therefore $a=b=1$.

It is correct.

The fact that the characteristic polynomial is equal to $(x-1)^2(x+1)^2$ shows that the dimension of the space is $4$ and that $\sigma(A) = \{-1,1\}$.

Now the fact that the minimal polynomial is equal to $(x-1)^2(x+1)^2$ shows that the largest blocks associated with $-1$ and $1$ are both of sizes $2 \times 2$.

Hence there is precisely one $2 \times 2$ block for $-1$, and one $2\times 2$ block for $-1$.