# Finding a non-diagonal matrix that can operate similar to a diagonal matrix

Suppose we have diagonal square($n \times n$) matrix $A$. (entries on diagonal are non-zero.)

1) Can there be non-diagonal square matrix $C$ that there exists non-zero vector $B$ with all entries non-negative of size ($1 \times n$) that satisfies $BC = BA$? And for any $A$, does $C$ always exist?

2) Continuing from 1), is there any such $C$ that $BC=BA$ all the time regardless of the choice of $B$? (Again $B$ has all entries non-negative and is not zero vector.) Again, for any $A$, does $C$ always exist?

Edit: Of course the case would be divided into two: when entries of $A$ on diagonal are same and the other. I would like to consider both cases.

## 1 Answer

1) $B C = B A$ means $C = A + T$ where $T$ is any matrix whose columns are all orthogonal to the row vector $B$. For example, take one nonzero column vector orthogonal to $B$ and use it for each row of $T$, and you'll get a non-diagonal matrix.

2) No, if $e_j C = e_j A$ for all $j$ (where $e_j$ is the row vector with $1$ in position $j$ and $0$ elsewhere) then $C=A$.