= Solution
By <diagonalizability>, $V$ has an <eigenbasis> $e_1,\ldots,e_m$ with $Te_j=\lambda_je_j$. Arrange the <basis> so that $\lambda_1=\cdots=\lambda_r=0$ and the remaining <eigenvalues> are nonzero. If there are no zero or no nonzero <eigenvalues>, the corresponding <linear span> below is simply the zero <vector subspace>.
For $v=\sum_j a_je_j$, the equation $Tv=0$ holds precisely when $a_j=0$ for every $j>r$. Thus
$$
\ker T=\operatorname{span}\{e_1,\ldots,e_r\}.
$$
Every output is in $\operatorname{span}\{e_{r+1},\ldots,e_m\}$, and conversely $e_j=T(\lambda_j^{-1}e_j)$ for each $j>r$. Therefore
$$
\operatorname{im}T=\operatorname{span}\{e_{r+1},\ldots,e_m\}.
$$
These two disjoint parts of a <basis> span $V$ and their <linear spans> intersect only at zero. This is exactly the <direct sum of vector spaces> assertion
$$
\boxed{V=\ker T\oplus\operatorname{im}T.}
$$
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