Let $V$ be the space of $n\times n$ matrices over a field $F$. Let $A$ be a fixed matrix in $V$. Let $T$ and be linear operators on $V$ defined by
$T(B)=AB$
I was trying to approach it in a way so that we don't have to use minimal polynomial.
To show that $T$ is diagonalizable we need to show that there exists $n^2$ distinct matrices $\{B_{11},\cdots B_{nn} \}$ such that $T(B_{ij}) = \lambda_j B_{ij}$ .
We see that $A$ is diagonalizable means there exists $n$ distinct vectors $\{v_1,\cdots , v_n\}$ such that $T(v_i) = c_j.v_i$.
Let $\{e_1,\cdots e_n\}$ be the standard basis of $V$ over $\mathbb{F}$ then I was trying to construct $B_{ij}$ as $B_{ij}(e_i) = v_j$ and $0$ otherwise (for all $1 \le i \le n$).
Then $T(B_{ij}) = AB_{ij}$ .
Now $A.B_{ij}(e_i) = A(e_i) = c_k.v_j = c_k.B_{ij}(e_j)$ .
Then $T(B_{ij})= c_k.B_{ij}$.
Can someone help me out with this approach?