How do I know that an inverse of a matrix has the same type of Jordan canonical form Let $A$ be an invertible matrix in $M_n(\mathbb{C})$.
How do I prove that $A^{-1}$ has the same block structure in its Jordan canonical form as $A$ does?
For each $x\in \mathbb{C}^n, A(x)=\lambda x$ iff $A^{-1}(x)=1/\lambda x$. Hence, they have exactly the same number of distinct eigenvalues.
So WLOG, we only need to check for the case $A$ has exactly one eigenvalue $\lambda$.
Consider a cycle of generalized eigenvectors of $A$ corresponding to $\lambda$. That is, $\{v_1,...,v_k\}$. However, this does not need to be a cycle of generalized eigenvectors of $A^{-1}$ corresponding to $1/\lambda$. So I don't have a clue here. How do I prove it?
 A: Suppose the Jordan canonical form of $A$ is given by the block diagonal form
$J = \operatorname{diag} (J_1,...,J_m)$, where each $J_k$ has the form
$J_k = \begin{bmatrix}
                 \lambda_k & 1 & 0 & \cdots & 0 \\
                  0 & \lambda_k & 1 & \cdots & 0 \\
                  0 & 0 & \lambda_k & \cdots & 0 \\
                  \vdots & & & & \vdots \\
                  0 & 0 & 0 & \cdots & \lambda_k
              \end{bmatrix}$.
Since $A$ is invertible, we see that each $J_k$ is invertible.
Then we see that
$J^{-1} = \operatorname{diag} (J_1^{-1},...,J_m^{-1})$.
Hence we only need to establish that the Jordan canonical form of $J_k^{-1}$ has the same form as $J_k$.
One simple way to do this is to note that
$J_k v = \lambda_k v_k$ iff $J_k^{-1} v = {1 \over \lambda_k} v$,
and so $\ker (J_k -\lambda_k I) = \ker (J_k^{-1}-{1 \over \lambda_k} I)$. In particular, $\dim \ker (J_k^{-1}-{1 \over \lambda_k} I) = 1$
and so the Jordan Canonical form of $J_k^{-1}$ is
$\begin{bmatrix}
                 {1 \over \lambda_k} & 1 & 0 & \cdots & 0 \\
                  0 & {1 \over \lambda_k} & 1 & \cdots & 0 \\
                  0 & 0 & {1 \over \lambda_k} & \cdots & 0 \\
                  \vdots & & & & \vdots \\
                  0 & 0 & 0 & \cdots & {1 \over \lambda_k}
              \end{bmatrix}$ (with the same dimensions as $J_k$,
of course).
