AandBare non-negative nxn matrices.

Show that eigenvalues of(I − A)and^{-1}BB(I − A)are the same.^{-1}

Thanks!

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- Mar 23rd 2013, 10:08 AMgotmejerryHow to show that the eigenvalues are the same?
**A**and**B**are non-negative nxn matrices.

Show that eigenvalues of**(I − A)**and^{-1}B**B****(I − A)**are the same.^{-1}

Thanks! - Mar 23rd 2013, 05:00 PMchiroRe: How to show that the eigenvalues are the same?
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. - Mar 23rd 2013, 05:07 PMButterflyMRe: How to show that the eigenvalues are 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. - Mar 24th 2013, 02:57 AMILikeSerenaRe: How to show that the eigenvalues are the same?
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)**=t^{-1}B v**v**.

Left multiply with**B**.

**B(I − A)**=^{-1}B v**B**t**v**

Let**w**=**Bv**. Then

**B(I − A)**=t^{-1}w**w**

So t is also an eigenvalue of**B(I − A)**, but only if^{-1}**Bv**=**w**is not the null vector.