Eigenvalues for an expression

Question

While I was working I found a problem in which I encountered $$\det(A+xB)$$ where $A, B \in M_n(\mathbb{C}), x \in \mathbb{R}$. If we denote by $a_1, a_2, ..., a_n$ the eigenvalues of $A$ and by $b_1, b_2, ..., b_n$ the eigenvalues of $B$, it is known, for example, that $\det(A)=a_1a_2...a_n$. My question is:

Is it true that the eigenvalues of $A+xB$ are $(a_1+xb_1), (a_2+xb_2),...,(a_n+xb_n)$? Why? Does it hold for any kind of expression involving $2$ or more matrices?

For example, Lord Shark the Unknown used here that "only possible eigenvalues of $I+A-A^2$ are $1+0-0^2=1$ and $1+1-1^2=1$"

Answer 1

This is not true in general, and it is very easy to write a counterexample. You should do this as an exercise.

However, if $A$ and $B$ have the same eigenspaces, associated with the corresponding indexes, then your formula holds. This is also easy to show. The converse case also holds because of your variable $x$.

Answer 2

Let $A = \begin{bmatrix} 2 & 1 \\ 1 & -2\end{bmatrix}$, then the trace is $0$ with determinant $-5$, hence the eigenvalues are $\sqrt5$ and $-\sqrt5$.

Let $B= \begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}$, the trace is $0$ with determinant $1$, hence the eigenvalues are $i$ and $-i$.

If we add them together, we have $A+B=\begin{bmatrix} 2 & 2 \\ 0 &-2\end{bmatrix}$ with eigenvaluces $2$ and $-2$ which cannot be written as the sum of the eigenvalues of the eigenvalues of $A$ and $B$.