Jordan Cell with rectangular matrices

Question

Suppose $J(\lambda_{1},k_{1})$ and $J(\lambda_{2},k_{2})$ are two Jordan cells and suppose that $C$ is a $k_{1}\times k_{2}$ matrix such that $ J(\lambda_{1},k_{1})C=CJ(\lambda_{2},k_{2}).$ Show that $C=0$ if $\lambda_{1}\neq\lambda_{2}$

Answer 1

A little hint:

Note that we can write $$ J(\lambda,k) = J(0,k) + \lambda I $$ We can also rewrite your equality as $$ J(\lambda_1,k_1)C - CJ(\lambda_2,k_2) = 0 \implies\\ J(0,k_1)C + \lambda_1 C - CJ(0,k_2) - \lambda_2 C = 0 \implies\\ J(0,k_1)C - CJ(0,k_2) + (\lambda_1 - \lambda_2)C = 0 $$ What you need to show, then, is that $J(0,k_1)C - CJ(0,k_2)$ can never be a non-zero multiple of $C$.