# 8 Exercises

## 8.1 II.C.1

Let $$E$$, $$A \in \mathbb C^{n,n}$$ satisfy $$EA=AE$$. Then $$\ker E \cap \ker A = \{0\}$$ implies that $$(E, A)$$ is regular.

Assume that $$\ker A \neq \{0\}$$ is of dimension $$k\geq 1$$. The case that $$k=0$$ is trivial, since $$\lambda E - A$$ is regular for $$\lambda = 0$$. Let $$V_0$$ be the matrix whose columns span $$\ker A$$ and let $$V_\perp$$ be the matrix that consists of all eigenvectors of $$A$$ that are associated with the nonzero eigenvalues.

It holds that,

$AV_0 = 0 \quad \text{and} \quad AV_\perp = V_\perp L_\perp$

with an $$L_\perp \in C^{n-k,n-k}$$ which is invertible. This a consequence of $$V_\perp$$ spanning the $$A$$-invariant subspaces with respect to the nonzero eigenvalues.

Because of $$ABV_0=BAV_0=0$$, it follows that $$\operatorname{span}BV_0 \subset \ker A = \operatorname{span}V_0$$, i.e., $$V_0$$ is a $$B$$-invariant subspace which means that there is a $$K_0\in \mathbb C^{k,k}$$ such that $$BV_0 =V_0K_0$$.

Moreover, because of $$\ker E \cap \ker A = \{0\}$$, the matrix $$K_0$$ has no zero eigenvalues. In fact $$K_0$$ has the same eigenvalues as $$B':=B\bigr|_{V_0}\colon V_0 \to V_0$$, and if $$B'$$ had a zero eigenvalue this would mean that the associated eigenvector would be in $$V_0$$ and, thus, in the kernel of $$A$$.

Moreover, since $$ABV_\perp=BAV_\perp=BVL_\perp$$ meaning that $$BV_\perp$$ is in the $$A$$-invariant subspace related to the nonzero eigenvalues of $$A$$, i.e., $$BV_\perp \subset V_\perp$$, it follows that $$V_\perp$$ is a $$B$$-invariant subspace and, thus, $$BV_\perp = V_\perp K_\perp$$ for some matrix $$K_\perp \in \mathbb C^{n-k,n-k}$$.

With $$V:=[V_0 |V_\perp]$$ and the observation that $$V$$ is invertible, since its columns span all of $$\mathbb C^n = \operatorname{span}V_0 \oplus \operatorname{span}V_\perp$$, it follows that $\begin{equation*} \begin{split} \lambda E - A & = (\lambda E - A)VV^{-1} = (\lambda E [V_0 |V_\perp]- A[V_0 |V_\perp])V^{-1} \\ & = ([V_0 K_0 |V_\perp K_\perp]\lambda - [0 |V_\perp L_\perp])V^{-1} \\ & = [V_0 |V_\perp ] \begin{bmatrix} \lambda K_0 & \\ & \lambda K_\perp - L_\perp \end{bmatrix} V^{-1} \end{split} \end{equation*}$ and that $\det (\lambda E - A) = \det (\lambda K_0) \det(\lambda K_\perp - L_\perp)$ is not identically zero, since $$K_0$$ and $$L_\perp$$ are invertible.