A proof about vector space and diagonalization

Question

Context: I am studying a proof about diagonalizable and nilpotent matrix. I don't understand one step of the proof, but I think I am missing something obvious.

Let $E$ be a $k$-vector space of dimension $n$, and $B=(e_1,\ldots,e_n)$ be a basis of $E$.

Let's define an endomorphism $u$, such that $u(e_i)=\lambda_i e_i$ for all $1\leqslant i\leqslant n$.

And let's define another endomorphism $E_{ij}$ such that $E_{ij}(e_k)=\delta_{jk}e_i$ where $\delta$ is Kronecker's symbol.

Then, why is

$$ \mathrm{ad}_u(E_{ij})=(\lambda_i-\lambda_j)E_{ij}$$

where $\mathrm{ad}_u(v)=uv-vu$ for all $v\in \mathrm{End}(E)$?

I think I just need to manipulate the definitions, but I can not figure it out...

Answer 1

Let us calculate the action of $\operatorname{ad}_u(E_{ij})$ on $e_k$:

$$ \operatorname{ad}_u(E_{ij})(e_k) = u(E_{ij}(e_k)) - E_{ij}(u(e_k)) = u(\delta_{jk} e_i) - E_{ij}(\lambda_k e_k) = \delta_{jk} \lambda_i e_i - \lambda_k \delta_{jk} e_i = \delta_{jk}(\lambda_i - \lambda_k) e_i.$$

Now let us calculate the action of $(\lambda_i - \lambda_j) E_{ij}$ on $e_k$:

$$ ((\lambda_i - \lambda_j) E_{ij})(e_k) = (\lambda_i - \lambda_j) \delta_{jk} e_i = (\lambda_i - \lambda_k) \delta_{jk} e_i. $$

In the last equality, we used that if $j \neq k$ the both expressions are zero and if $j = k$ then $\lambda_i - \lambda_j = \lambda_i - \lambda_k$. Hence, $\operatorname{ad}_u(E_{ij}) = (\lambda_i - \lambda_j) E_{ij}$ because they agree on a basis.