Under what conditions is $\det(\alpha A + B)=0$ true for a $\alpha \in \Bbb R$?

Question

Given two real matrices $A$ and $B$ where $A$ is symmetric ($A^T=A$), can one deduce some condition(s) under which

$$\det(\alpha A + B)=0$$

is satisfied only if $\alpha \in \Bbb R$? In my case, $ A$ and $ B$ are $8 \times 8$ matrices.

Answer 1

As pointed out by @whpowell96,

If $A$ and $B$ share a common eigenvector $v$ with eigenvalues $\lambda_A\neq 0$ and $\lambda_B\in\mathbb{R}$ respectively, then $\alpha=-\frac{\lambda_B}{\lambda_A}$ is a real value such that $\det(\alpha A+B)=0$.

This is not hard to prove. First, a real symmetric matrix is diagonalizable with real eigenvalues, so $\lambda_A$ and hence $\alpha$ are real. Also, if a matrix has a zero eigenvalue, then it is not invertible and has zero determinant. We need only show that $\alpha A+B$ has a zero eigenvalue. Note that $(\alpha A+B)v=(\alpha\lambda_A+\lambda_B)v=0v$, so we have shown $\det(\alpha A+B)=0$.

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Note that such $\alpha$ may not always exist. Consider $$ A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},\ B = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} $$ Then $\det(\alpha A+B)=\alpha^2+1$, so no real $\alpha$ vanishes the determinant.