In his notes, my instructor began with the definition that
$T \in L(V)$ is diagonalizable $\leftrightarrow$ $V$ has a basis each vector of which is an eigenvector of $T$.
Let $\lambda_1, \lambda_2 … \lambda_m$ be the distinct eigenvalues of a diagonalizable $T$, and let $W_i = \ker(T-\lambda_iI)$. Then, he wrote, $ V = W_1 + W_2 … + W_m$ by the definition of a diagonalizable $T$. $V = W_1 \bigoplus W_2 … \bigoplus W_m$ by the fact that the eigenvalues are distinct.
How come the premise that $T$ has a basis of eigenvectors is not sufficient for $V = \bigoplus_i W_i$?
At first, I confused it with $V = \bigoplus_i \mathrm{span}\{v_i\}$, where $\{v_1, v_2 …v_m\}$ is an eigenbasis of $V$.
But now I know that they are not the same thing. Suppose that $V = \mathbb R^n$ and $T = I$; $\mathrm{span}\{e_1\} \subsetneqq \ker(T-I)$.
Best Answer
To have $V = \bigoplus_i W_i$ requires not just knowing something of $T$, but also of what the $W_i$ are. If one would list some of the eigenvalues more than once, for instance this could happen if one would list the roots of the characteristic polynomials with their multiplicities (as is often done in other circumstances), then some of the spaces $W_i$ would be equal, and in that case the sum cannot be direct. So "by the fact that the eigenvalues are distinct" is just there to remind you that this aspect of the set-up is being used to conclude the directness of the sum. (It is not really a hypothesis, because in any situation one can decide to list the eigenvalues without repetitions, but it is important that this was actually done here.) It is a theorem (apprently being used here) that any sum of eigenspaces for distinct eigenvalues is a direct sum.
Note that if you would be talking of the direct sum of the spans of eigenvectors in a basis, the sum would have $n$ terms, not $m$. In fact some eigenspaces $W_i$ might themselves be decomposed as the direct sum of $1$-dimensional spaces spanned by eigenvectors. If this happens, then there are in fact infinitely many ways in which $W_i$ can be so decomposed, because one can choose any basis of $W_i$ as part of a basis of eigenvectors. Also note that it cannot happen that the sum of the dimensions of the eigenspaces $W_i$ is more than$~n$ (by the cited theorem of directness of their sum), so a basis of eigenvector must consist of a union of bases of the eigenspaces, there is no room to choose fewer vectors in an eigenspace than its dimension. The theorem does not imply that the sum of the dimensions of the eigenspaces is always equal to$~n$, it could be less if (and only if) $T$ is not diagonalisable.