How do the eigenpairs of a matrix change after some matrix multiplication? I am interested to know if there is a way to find the eigenpairs of matrix $AT$ or $TA$ from $A$, where $A$ is an arbitrary matrix and $T$ is some transform matrix.
Edit
$$AV = \lambda V$$
$$(TA)V_1 = \lambda_1 V_1$$
$$(AT)V_2 = \lambda_2 V_2$$
Is there a relation between $V,V_1,V_2$ using $T$?
 A: Matrix multiplication essentially corresponds to composing linear transformations on some vector space. For this reason, there is no reason to expect that there would be a nice relationship between eigenvectors and eigenvalues after the transformation. For example, let $A,T\colon\mathbb R^2\to\mathbb R^2$ where $$A=\begin{bmatrix}1&0\\ 0&1\end{bmatrix}, T=R_\theta=\begin{bmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$$
so $A$ is the identity, and $R_\theta$ is rotation counterclockwise by angle $\theta$. Now, if $\theta$ is a generic angle like, say, $\pi/3$, then $T$ has no eigenvectors, because everything is rotated and nothing is fixed in direction. This shows we cannot have some kind of relation between the eigenvectors of $A$ and $TA$, nor between $A$ and $AT$, because eigenvectors might not even exist in the first place!
On the other hand, there is something meaningful to be said about $TAT^{-1}$. Notice that, algebraically, $TAT^{-1}(Tv)=TAv = T\lambda v = \lambda(Tv)$ given any eigenvector $Av=\lambda v$. Thus the eigenvectors are transformed by $T$, and eigenvalues remain fixed. This is conceptually because any expression of the type $TAT^{-1}$ can be interpreted as a change of basis according to $T$, and $Tv$ is simply the expression for the vector in terms of the new coordinates. This holds whenever $T$ is invertible.
