Spectral ratio.

djimenez

Hello,

I found this one very nice.

1) If $F\in\mathcal M_{n\times m}$ with $m\neq n$ , show that $\rho(FF^T)=\rho(F^TF)$ , where $\rho(X)$ prepresent the spectral ratio of the square matrix $X$ (take it general, because in the first case the matrix is $n\times n$ and in the second $m\times m$ ).

2) If $A,B\in\mathcal M_{n\times m}$ , is it true that $\rho(AB^T)=\rho(A^TB)$ or $\rho(AB^T)=\rho(B^TA)$ ?

Best,

**Posted:** Thu Mar 24, 2005 9:44 am

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David Jimenez

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djimenez

Anyone? It is a nice linear algebra problem!

**Posted:** Wed May 04, 2005 10:00 am

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Peter

what is spectral ratio? I thought the set of eigenvalues but then the thing is false, so what is it? Smile

**Posted:** Thu May 05, 2005 3:05 pm

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djimenez

**Posted:** Thu May 05, 2005 3:33 pm

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Peter

Aren't both a trivial consequence from {eigenvals(AB)} union {0} = {eigenvals(BA)} union {0}? Blush

Or is that the theorem we have to prove? Smile

**Posted:** Fri May 06, 2005 2:58 am

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Boo!

djimenez

**Posted:** Fri May 06, 2005 8:19 am

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There is just one thing that is easier to get than to lose, WEIGHT
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Peter

Look: let k be an eigenvalue, x be an eigenvector:
ABx = kx, k<>0
BABx = Bkx = kBx
=> Bx is an eigenvector of BA, corresponding to eigenvalue k

Razz

**Posted:** Fri May 06, 2005 8:52 am

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jmerry

That's the simple version. If you don't have eigenvectors, use this.

Of course it is- just pad both matrices to square with zeros: $\begin{bmatrix}A&0\end{bmatrix}\cdot\begin{bmatrix}B\\0\end{bmatrix}= \begin{bmatrix}AB\end{bmatrix}$
and $\begin{bmatrix}B\\0\end{bmatrix}\cdot\begin{bmatrix}A&0\end{bmatrix}= \begin{bmatrix}BA&0\\0&0\end{bmatrix}$
(assuming WLOG that AB is larger than BA, and choosing the zero blocks to match dimensions).

This then reduces to the case for square matrices, in which $\det A\det B\det(\lambda I-AB)=\det(B(\lambda I-AB)A)=\det(\lambda BA-BABA)=\det BA\det (\lambda I-BA)=\det B\det A\det(\lamb...$ . That says that for invertible $A,B$ the characteristic polynomial of $AB$ is the same as the characteristic polynomial of $BA$ . Since the coefficients of the characteristic polynomial are polynomials in the coefficients of $A$ and $B$ which agree almost everywhere, they must be identical polynomials. (The coefficients of this polynomial are rational; work over $\mathbb{C}$ to make sure this works. Since the coefficients depend only on dimension and not on the field, this then applies in a general field).

**Posted:** Fri May 06, 2005 8:55 am

Peter

**Posted:** Fri May 06, 2005 8:58 am

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Boo!

jmerry

What if there aren't any eigenvalues at all? For all you know, I'm working over the rationals or some finite field. The spectral radius question assumes $\mathbb{C}$ , but the fact that the characteristic polynomial of $AB$ is a power of $x$ times the characteristic polynomial of $BA$ doesn't.

**Posted:** Fri May 06, 2005 9:07 am

djimenez

**Posted:** Fri May 06, 2005 10:19 am

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There is just one thing that is easier to get than to lose, WEIGHT
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Peter

**Posted:** Fri May 06, 2005 11:32 am

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Boo!

jmerry

To make those unambiguous eigenvalues, pad the matrices out to square as in my argument.

**Posted:** Fri May 06, 2005 12:11 pm