# 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.