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Assume matrix $A$ is diagonalizable and has positive real eigenvalues. I just don't understand how I am supposed to know how the eigenvalues interact without knowing the exact matrices specified in the problem. My textbook doesn't mention anything about this and I'm starting to get confused.

anomaly
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  • To see that the eigenvalues of a sum are tricky, try experimenting with e.g. adding $2 \times 2$ matrices where one is upper triangular and one is lower triangular. The individual matrices have easy to see eigenvalues, but the sum really doesn't. – Jakob Streipel Dec 10 '21 at 18:46

2 Answers2

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In general saying something about the eigenvalues of the sum of two matrices is really hard, even if they are diagonalizable.

However: if they are simultaneously diagonalizable it is a different story.

Suppose that for matrices $A$ and $C$ we have that both $PAP^{-1}$ and $PCP^{-1}$ are diagonal using the same matrix $P$ then it is easy:

The eigenvalues of $A + C$ equal the eigenvalues of $P(A + C)P^{-1}$ and this matrix in turn equals $PAP^{-1} + PBP^{-1}$ and hence is the sum of two diagonal matrices. Computing these eigenvalues is easy.

So the hard part is: how do we know, for given $A$ and $C$ if such a $P$ exist. There is a really nice theorem that states that if $A$ and $C$ are both diagonalizable and they commute (that is: $AC = CA$) then they are simultaniously diagonalizable.

In your case with $C = A^{-1}$ this condition is satisfied.

EDIT: this older question might shed some light on the 'really nice theorem': Simultaneous Diagonalizability of Multiple Commuting Matrices

Vincent
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In general if the eigenvalues of two matrices have no bearing on the eigenvalues of their sum. The exception would be the case where $v$ is an eigenvector of both matrices.

In the example....

Suppose $v$ is an eigenvector of $A$ with eigenvalue $\lambda$ then $v$ will be an eigenvector of $A^{-1}$ with eigenvalue $\frac {1}{\lambda}$

$Bv = Av + A^{-1}v - 2Iv = \lambda v + \frac {1}{\lambda}v - 2v$

And $v$ is an eigenvector of $B$ with eigenvalue $\lambda + \frac {1}{\lambda} - 2$

user317176
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  • This was the exact example I needed to figure out the rest of the problem above. You and vincent's answers are literally perfect! Thank you –  Dec 10 '21 at 20:29