A and B are non-negative nxn matrices.
Show that eigenvalues of (I − A)^{-1}B and B(I − A)^{-1} are the same.
Thanks!
Hey gotmejerry.
First consider the determinant of both expressions which is the product of the eigen-values. If all eigen-values are different then you should have linear independence which implies the matrices are non-singular.
If this is the case and you have three different non-zero eigenvalues with the same determinant for both expressions, then the eigenvalues have to be the same.
After a few wrong turns, I think I got now right for you... "If two matrices have the same rank, trace and determinant, they have the same eigenvalues"
So all you need to do is use a bit matrix algebra and in this case:
Let's get the determinant first:
$\displaystyle det[(I-A)^{-1}B] = det[B]/det[I-A]$
Note that this is the same as $\displaystyle det[B(I-A)^{-1}= det[B]/det[I-A]$
Therefore both matrices have the same determinant.
The trace has the commuative property, that is trace(AB)=trace(BA),
Therefore both matrices have also the same trace.
Both statements imply that $\displaystyle (I-A)^{-1}B$ and $\displaystyle B(I-A)^{-1}$ have the same n non-negative eigenvalues.
Note, however, both matrices might not have the same eigenvectors, or in other words, the eigenvalues must not be necessarily in the same order.
I'm afraid this is only true of 2x2 matrices.
For 3x3 matrices this gives you 2 equations with 3 unknowns which has more than 1 solution.
Suppose t is an eigenvalue of (I − A)^{-1}B .
Then there is a v such that (I − A)^{-1}B v=tv.
Left multiply with B.
B(I − A)^{-1}B v=Btv
Let w=Bv. Then
B(I − A)^{-1}w=tw
So t is also an eigenvalue of B(I − A)^{-1}, but only if Bv=w is not the null vector.