$T(A)=BA$ implies geometric multiplicity of every eigenvalue of $T$ is $\ge n$.

Question

Let $T: \mathbb C ^{n \times n} \to \mathbb C ^{n \times n}$ defined by $T(A)=BA$.

$A,B \in \mathbb C ^{n \times n}$.

(I already proved that every eigenvalue of $T$ is eigenvalue of $B$, and vice versa.)

Prove:

• Geometric multiplicity of every eigenvalue of $T$ is $\ge n$.

Answer 1

Hint: Let $x$ be an eigenvector of $B$. Let $A$ be a matrix with zeros in every column except for the $j$th column, and take the $j$th column to be the vector $x$.