# Operators and eigenvalues

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?

## 1 Answer

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.