Is the inverse of a block matrix also a block matrix?

Question

If $N$ is a nilpotent matrix then $N^t$ and $N$ are similar. Use the jordan form and this to prove that a complex matrix is similar to transpose.

Let $N$ be a $k \times k$ nilpotent matrix such that $N^{K-1} \ne 0$ such that $N^k= 0$.Now we see that characteristic polynomial and minimal polynomial are same then there exists a $v \ne \{0\}$ such that $v$ is a cyclic vector of $N$.

Then the jordan block of the matrix is,

$\begin{bmatrix} 0 & 0 & 0 & \cdots & 0 & 0\\ 0 & 1 & 0 & \cdots & 0 & 0\\ 0 & 0 & 1 & \cdots & 0 & 0\\ \vdots \\ 0 & 0 & 0 & \cdots & 1 & 0\\ \end{bmatrix}$

Now $N^t$ also has the same minimal polynomial as thaat of $N$ then the characteristic polynomial and the minimal polynomial are same and we can find a cyclic vector $v' \ne \{0\}$ such that it has the above jordan form.

Hence $N^t$ and $N$ are similar.

Let $J$ be jordan form of the matrix $N$ then let

$J = \begin{bmatrix} J_1 & 0 & \cdots & 0 \\ 0 & J_2 & \cdots & 0 \\ \vdots \\ 0 & 0 & \cdots & J_r \\ \end{bmatrix}$

where $J_i$ are all jordan blocks.

Then $J^t = \begin{bmatrix} (J_1)^t & 0 & \cdots & 0 \\ 0 & (J_2)^t & \cdots & 0 \\ \vdots \\ 0 & 0 & \cdots & (J_r)^t \\ \end{bmatrix}$

We see that $J_i = \begin{bmatrix} c & 0 & \cdots & 0\\ 1 & c & \cdots & 0 \\ \vdots \\ 0 & 0 & \cdots & c \\ \end{bmatrix}$

Then $(J_i - cI)^k = 0$ and $((J_i)^t - cI)^k = 0$ . They are both nilpotent operators which have the same order so they are similar and $P_i(J_i - cI)P_i^{-1} = J_i^t - cI$ . Hence $P_i J_i .P_i^{-1} = (J_i)^t$.

So I thought if I multiply $J$ by the matrix $P = \begin{bmatrix} P_1 & 0 & \cdots & 0\\ 0 & P_2 & \cdots &0 \\ \vdots \\ 0 & 0 & \cdots & P_k\\ \end{bmatrix}$

if $P^{-1} = \begin{bmatrix} P_1^{-1} & 0 & \cdots & 0\\ 0 & P_2^{-1} & \cdots &0 \\ \vdots \\ 0 & 0 & \cdots & P_k^{-1}\\ \end{bmatrix}$ .Then $PJP^{-1} = J^t$

I am not sure whether what I claimed is correct or not. Some hints would be helpful..

Answer 1

A block diagonal matrix is invertible if and only if each block is invertible. Now form another block diagonal matrix that consists of the inverses of each block. Can you see the answer now?

Answer 2

The product of two matrices defined by blocks is the matrix formed with the products of the blocks. As for all $i$, you have $P_i J_i .P_i^{-1} = (J_i)^t$, the conclusion

$$P J .P^{-1} = J^t$$ is indeed correct.