= Solution
Since $T$ is <diagonalizable>, its <minimal polynomial> is $\prod_{j=1}^r(x-\lambda_j)$ with distinct <eigenvalues> $\lambda_j$ in the <field>. Apply the <primary decomposition theorem for linear maps>:
$$
\boxed{V=\bigoplus_{j=1}^r\ker(T-\lambda_jI)=\bigoplus_{\lambda\in\mathbb F}V_\lambda.}
$$
For a <scalar> that is not an <eigenvalue>, the associated <eigenspace> is zero, so the sum indexed by the whole <field> has only finitely many nonzero summands. The <eigenspaces> need not be one-dimensional.
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