Vector space$V$ is direct sum of eigenspaces of $\phi$ when $\phi $ is diagonalisable

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This seems a simple statement but if $\phi \in \text{End}_K (V)$ is diagonalisable then I am trying to show that $$V=V_{\lambda_1} \oplus \cdots \oplus V_{\lambda_n} $$ where $\lambda_1 ,\ldots, \lambda_n$ are the distinct eigenvalues (though we may have $r < \dim (V) ) $.

I know that in general we have a direct sum into generalised eigenspaces but this above is a stronger result when $\phi $ is diagonalisable.