How does composition affect eigendecomposition? 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.
 A: 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 $AT$.   
A: 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.
A: 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 $\nu_1,\nu_2$.
