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
Best Answer
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.