3
$\begingroup$

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
$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ how does this mean that the eigenvalues are real in this case? $\endgroup$ – user66906 May 16 '14 at 13:06
  • $\begingroup$ @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. $\endgroup$ – egreg May 16 '14 at 13:16
  • $\begingroup$ yes, but I checked these matrices for let's say 100 cases and there always was a real eigenvalue. Now I was wondering whether there is a mathematical reason for this. Of course, it can only be a coincidence, but if there is a property of the matrices so that there needs to be a real eigenvalue, then this would answer my question. $\endgroup$ – user66906 May 16 '14 at 13:20
  • 1
    $\begingroup$ @Lipschitz You have been lucky; note, though, that odd ordered matrices always have a real eigenvalue. $\endgroup$ – egreg May 16 '14 at 13:48
  • $\begingroup$ see, your last point was actually good. so maybe something similar can be said for the even cases too. $\endgroup$ – user66906 May 16 '14 at 14:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy