The question is:
There is a basis of $V$ consisting entirely of eigenvectors of $T$ $\implies$ T is diagonalizable, where $T$ be a linear operator over $V$ with a basis $B$.
I've looked at the solution to this question, however, I don't understand part of it.
My solution has $$[Tv_i]_B = λ_i[v_i]_B = λ_ie_i,$$ where $e_i$ is the standard basis matrix. I understand the first equality, but not the second equality. Could someone please explain it a little bit?
Thanks.