If A and B are invertible square matrices, there exists

[AntiElephant]

Homework Statement

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.

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 kSo -\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?

 

[HallsofIvy]

Homework Statement

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).

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

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 kSo -\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?

 

[AntiElephant]

My bad...early morning. Edited.

 

[Dick]

I guess it is not necessarily true when k is not algebraically closed?

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?

 

[D H]

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.

 

[Dick]

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.

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.