# If A and B are invertible square matrices, there exists

1. Apr 2, 2014

### AntiElephant

1. The problem statement, all variables and given/known data

If A and B are invertible matrices over an algebraically closed field $k$, show there exists $\lambda \in k$ such that $det(\lambda A + B) = 0$.

3. The attempt at a solution

Can anyone agree with the following short proof? I tried looking online for a confirmation, but I wasn't exactly sure how to search exactly for this problem.

Suppose $det(\lambda A + B) \neq 0 ~~\forall \lambda \in k$

Since $det(A^{-1}) \neq 0$, then

$det(A^{-1}(\lambda A + B)) = det(\lambda I + A^{-1}B) = det(A^{-1}B - (-\lambda)I) \neq 0 ~~\forall \lambda \in k$

So $-\lambda$ is not an eigenvalue for $A^{-1}B$ for all $-\lambda \in k$. i.e. $\lambda$ is not an eigenvalue of $A^{-1}B$ for all $\lambda \in k$. Since $k$ is algebraically closed this cannot be the case.

Is this okay? I guess it is not necessarily true when $k$ is not algebraically closed?

Last edited: Apr 2, 2014
2. Apr 2, 2014

### HallsofIvy

Staff Emeritus
??? What are you asked to show about $det(\lambda A+ B)$? You have a subject with no verb!

3. Apr 2, 2014

4. Apr 2, 2014

### Dick

You bet it can fail. Take A=[[0,1],[1,0]] and B=[[2,0],[0,1]]. What goes wrong if k is the field of rationals?

5. Apr 2, 2014

### D H

Staff Emeritus
I didn't see that last bit about what happens when k is not algebraically closed. It can also fail with the reals. Just construct your matrices so the eigenvalues of A-1B are complex.

6. Apr 2, 2014

### Dick

Sure. But that's because the reals aren't algebraically closed. The det gives you a real polynomial. If it has no real roots, you are out of luck.