# Generalized eigenvalue problem; why do real eigenvalues exist?

Under this text, you can see two pairs of matrices $(A,C)$. I am currently solving the generalized eigenvalue $(A-\lambda C)v=0$ for several pairs of $(A,C)$ and found out that they also have real eigenvalues. I was therefore wondering, is there any property that these two matrices have(and which I could check on my other problems), so that it is a priori clear, that there will be a REAL eigenvalue to the generalized eigenvalue problem?

My second question would be: Is there anything that would tell us that there is a corresponding real eigenvector to the real eigenvalue?

A =

-0.5000    0.5000
5.5000    2.5000


C =

0.5000    0.5000
7.5000    3.5000


A =

-3.0000    0.5000         0
2.7500    3.2500    1.7500
9.0000    8.5000    4.0000


C =

-2.0000    0.5000         0
4.0000    4.0000    2.0000
11.0000   10.5000    5.0000


Basically, we are talking here about the eigenvalues of the matrix $AC^{-1}$

1.3333    0.8333   -0.3333
0.0833    2.3333   -0.5833
0.0000   -0.5000    1.0000


and

2.7500   -0.2500
-0.2500    0.7500


okay, the last case is easy, cause the matrix is symmetric.

In order to give you more information I also write down another $AC^{-1}$ matrix

Columns 1 through 3

1.2396    0.4687    1.5937
0.2648    2.2789    5.1817
0.2314    1.0613    6.0752
0.1076   -0.4062   -2.2813


Column 4

-0.8021
-2.5835
-2.5469
1.9618


If $C$ is invertible, then $$\det(A-\lambda C)=\det(AC^{-1}-\lambda I)C)=\det(AC^{-1}-\lambda I)\det C$$ so the eigenvalues of the pair are exactly the eigenvalues of $AC^{-1}$.
If $C$ is not invertible, then existence of eigenvalues of the pair is not guaranteed: for instance, if $A$ is invertible and $C=0$, then $(A-\lambda C)v=0$ implies $v=0$.
Notice that you are not going to get real eigenvalues in all cases. Just consider $C=I$ and $A$ without real eigenvalues.
• @Lipschitz They need not be real, of course. If $C$ is invertible and $AC^{-1}$ has no real eigenvalue, then you can't get real eigenvalues of the pair. – egreg May 16 '14 at 13:16