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Let $m\times m$ matrix B has eigenvalues $\beta_1,\beta_2, ... \beta_n$ and eigenvectors $b_1, b_2, ... b_n$. I need to find eigenvalues and eigenvectors for operator $$R: Mat_{n\times m} \rightarrow Mat_{n\times m}, R(x) = XB^T. $$

As I understand $XB^T = \beta X$ where $X$ is a left eigenvector and $\beta$ is a left eigenvalue.

Then $(XB^T)^T=BX^T$ and operator $$R^T(X): Mat_{m\times n} \rightarrow Mat_{m\times n}, R^T(X)=BX^T $$ After that I said that $BX^T=\beta'X^T$ and by using properties of determinants $det|B^T-\beta|=det|B-\beta|^T=det|B-\beta|=det|B-\beta'| \Rightarrow \beta = \beta'$.

Also I know eigenvalues and eigenvectors for $R^T(X)=BX^T$, they equal: $\beta_i \ and \ b_i. $

After all these considerations I found that $X^T and\ X$ are orthogonal and their product commutes with a diagonal matrix of eigenvalues, but I don't know how to find the explicit form of X.

Could you help please?

Could you correct my solving please?

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Your operator $R^T$ is much more complicated to study than your operator $R$.

For $R$, see

https://en.wikipedia.org/wiki/Kronecker_product

Here $R=I\otimes B$ if we stack the matrices row by row.

The eigenvalues of $R$ are $\beta_1,\cdots,\beta_m$, each $n$ times. If $b_1,\cdots,b_m$ are linearly independent, then $B$ is diagonalizable and $R$ also. It is easy to obtain the associated eigenvectors.

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