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?
Best Answer
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).