Fill in the details of the following argument to show that every \(n \times n\) matrix \(A\) with entries in \(\mathbb{H}\) has an eigenvalue in \(\mathbb{H}\). For \(t \in [0, 1]\) and \(\lambda \in S^3 \subset \mathbb{H}\), consider the matrix \(t\lambda I + (1-t)A\). If \(A\) has no eigenvalues, this is invertible for all \(t\). Thus the map \(S^3 \to GL_n(\mathbb{H})\), \(\lambda \mapsto \lambda I\), is nullhomotopic. But by the preceding problem and Exercise 10(b) in §3.C, this map represents \(n\) times a generator of \(\pi_3 GL_n(\mathbb{H})\).
Solution
Assume, for contradiction, that \(A\in M_n(\mathbb H)\) has no (left) eigenvalue. In particular \(A\) is invertible. For \[ t\in[0,1],\qquad \lambda\in S^3\subset\mathbb H, \] set \[ H(t,\lambda)=t\lambda I+(1-t)A. \] If \(0\le t<1\) and \(H(t,\lambda)\) were singular, there would be \(v\ne0\) with \[ Av=-\frac{t}{1-t}\lambda v, \] so \(-\frac{t}{1-t}\lambda\) would be an eigenvalue of \(A\), contrary to hypothesis. For \(t=1\), \(H(1,\lambda)=\lambda I\) is invertible. Thus \[ H:[0,1]\times S^3\to GL_n(\mathbb H) \] is a homotopy from the constant map \(\lambda\mapsto A\) to \[ \lambda\longmapsto\lambda I. \] Hence the latter map would be nullhomotopic.
But \(GL_n(\mathbb H)\) deformation retracts onto \(Sp(n)\), and the preceding exercise homotopes \(\lambda I=\operatorname{diag}(\lambda,\ldots,\lambda)\) to \[ \operatorname{diag}(\lambda^n,1,\ldots,1). \] By Exercise 10(b) of §3.C, the power map \(S^3\to S^3\), \(\lambda\mapsto\lambda^n\), represents \(n\) times a generator of \(\pi_3(S^3)\). The inclusion \(Sp(1)\hookrightarrow Sp(n)\) induces an isomorphism on \(\pi_3\), so \(\lambda\mapsto\lambda I\) represents \[ n\ne0\in\pi_3(GL_n(\mathbb H))\cong\mathbb Z. \] This contradicts nullhomotopy. Therefore every quaternionic matrix has an eigenvalue: \[ \boxed{\exists\lambda\in\mathbb H,\ v\ne0\text{ with }Av=\lambda v.} \]