An endomorphism is diagonalisable exactly when $V={⨁}_{i<r}{E}_{{\lambda }_i}(T)$ for some finite list of distinct scalars ${\lambda }_i$

Suppose $T$ is diagonalisable and fix an eigenbasis. Only finitely many eigenvalues occur among its finitely many vectors; group the basis vectors by those distinct eigenvalues. If $v\in E_{\lambda }(T)$ is expanded in the eigenbasis, comparing the coefficients in $Tv=\lambda v$ forces every coefficient at a basis vector with eigenvalue different from $\lambda$ to vanish. Thus the group carrying $\lambda$ is a basis of $E_{\lambda }(T)$. The groups exhaust the eigenbasis, so the eigenspaces sum to $V$; and a nonzero $v\in E_{\lambda }(T)$ lying in the sum of the other eigenspaces would have one eigenbasis expansion supported on the $\lambda$-group and another supported off it, contradicting uniqueness of coordinates in a basis. Both conditions of [L3] therefore hold.

Conversely, suppose the displayed direct sum holds. Choose a basis of each eigenspace using [L4] and concatenate the finite lists. The first condition of [L3] makes the concatenation spanning. It is independent: in a vanishing combination, group the terms by eigenspace, so each group sums to a vector of its $E_{\lambda }(T)$ and one such vector equals minus the sum of the others; the second condition of [L3] forces every group to sum to $0_V$, and independence inside each chosen basis then kills every coefficient. So it is a basis of eigenvectors and [L1] makes $T$ diagonalisable.

The two constructions prove both implications. When $V=0$, the empty basis and empty direct sum satisfy them.