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What relationship is there between the eigenvalues and vectors of linear operator $T$ and the composition $A T$ or $T A$? I'm also interested in analogous results for SVD.

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The case of $A$ invertible, $B$ arbitrary, and $T = A^{-1} B$ or $BA^{-1}$ might be of interest in refining the question. –  leslie townes May 10 '12 at 5:45

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Denote the eigenvalues of $T,A,TA$ by $\lambda_1,\lambda_2, \ldots, \lambda_n, \mu_1,\mu_2, \ldots, \mu_n, \nu_1,\nu_2, \ldots, \nu_n$, respectively. What relations exist between them ? There are several easy constraints :

(1) We must have $\prod_{k=1}^{n} \lambda_k\mu_k=\prod_{k=1}^{n}\nu_k$ because ${\sf det} (TA)={\sf det} (T) {\sf det} (A)$.

(2) We must have $k_{\nu} \geq {\sf max }(k_{\lambda},k_{\mu})$, where we denote by $k_{\lambda},(k_{\mu},k_{\nu})$ the number of indices $i$ such that $\lambda_i$ (or $\mu_i,\nu_i$ respectively), is zero (this is because ${\sf rank}(AT) \leq {\sf min}({\sf rank}(A),{\sf rank}(T))$ and ${\sf rank}(T)=n-k_{\lambda}$).

(3) If one of $A,T$ or $AT$ is a homothety, then $\lbrace \nu_{k} \rbrace_{1 \leq k \leq n}=\lbrace \lambda_k\mu_k \rbrace_{1 \leq k \leq n}$.

I think there are no other constraints besides the three above.

I can actually prove this for a "generic enough" example : assume $n=2$, and $\lambda_1\mu_1\lambda_2\mu_2=\nu_1\nu_2$ (to ensure (1)), and $\lambda_1 \neq \lambda_2, \mu_1 \neq \mu_2, \nu_1 \neq \nu_2$ (to ensure (3)).

Then consider the two matrices

$$ P=\bigg( \begin{matrix} (\nu_1-\lambda_1\mu_2) & (\nu_1-\lambda_1\mu_1)\\ (\nu_1-\lambda_2\mu_2) & (\nu_1-\lambda_2\mu_1)\\ \end{matrix} \bigg), $$

$$ Q=\bigg( \begin{matrix} (\nu_1-\lambda_2\mu_1)(\nu_1-\lambda_2\mu_2) & -(\nu_1-\lambda_1\mu_1)(\nu_1-\lambda_1\mu_2) \\ (\nu_2-\lambda_2\mu_1)(\nu_1-\lambda_2\mu_2) & -(\nu_2-\lambda_1\mu_1)(\nu_1-\lambda_1\mu_2) \\ \end{matrix} \bigg) $$ We have $$ {\sf det}(P)=\nu_2(\lambda_2-\lambda_1)(\mu_2-\mu_1), $$

$$ {\sf det}(Q)=\mu_1(\lambda_2-\lambda_1)(\nu_2-\nu_1)(\nu_1-\lambda_1\mu_2)(\nu_1-\lambda_2\mu_2) $$ so that $P$ and $Q$ are both inversible whenever $\mu_1\neq 0,\nu_2\neq 0, \nu_1 \not\in \lbrace \lambda_1\mu_1, \lambda_1\mu_2 \rbrace$. If we put $$ L={\sf diag}(\lambda_1,\lambda_2), M={\sf diag}(\mu_1,\mu_2), N={\sf diag}(\nu_1,\nu_2) $$ then $QLPM=NQP$ (GP can check this), so that $LPMP^{-1}=Q^{-1}NQ$. If we put $T=L$, $A=PMP^{-1}$, then $T$ has eigenvalues $\lambda_1,\lambda_2$, $A$ has eigenvalues $\mu_1,\mu_2$, $TA$ has eigenvalues $\nu_1,\nu_2$.

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The eigenvalues of $AT$ and $TA$ are the same. $\det(AT) = \det(A)\det(T)$, so the product (counting multiplicities) of the eigenvalues of $AT$ is $\det(A)$ times the product for $T$. I don't think there's much else you can say. For example, consider $A = \pmatrix{0 & 1\cr 1 & 0\cr}$ and $T = \pmatrix{1 & t\cr 0 & 1\cr}$, $AT = \pmatrix{0 & 1\cr 1 & t\cr}$. The only eigenvalue of $T$ is $1$; for any nonzero $\lambda$ you can choose $t=\lambda - 1/\lambda$ so that $\lambda$ and $1/\lambda$ are the eigenvalues of $AT$.

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A short proof would be: Let $ABx=\lambda x$, and let $y=Bx$. Then $BAy=BA(Bx)=B(ABx)=B\lambda x=\lambda y$. That also shows the connection between the eigenvectors. –  Keivan May 23 '12 at 19:09
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Yes, assuming $\lambda \ne 0$. –  Robert Israel May 23 '12 at 19:48
    
Do you mean assuming $Bx \neq 0$? –  Keivan May 23 '12 at 19:53

Friedland has proved the following over the complex field:

If the principal minors of $A$ are not zero, then for every set of $n$ numbers $\lambda_1,\dots,\lambda_n$ there exist a diagonal matrix $B$ such that $BA$ has $\lambda_i$'s as eigenvalues.

Later Dias da Silva extended it to any arbitrary algebraically closed field.

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I guess you forgot a condition on $B$ in your statement, because as it is stated now it is trivial. –  Ewan Delanoy May 24 '12 at 5:56
    
Yes, it is diagonal. I'll add it. –  Keivan May 24 '12 at 15:36

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