= Solution
Take the <linear map>
$$
\boxed{T(x,y)=(y,0),\qquad [T]=\begin{pmatrix}0&1\\0&0\end{pmatrix}.}
$$
Its <kernel of a linear map> consists of all $(x,0)$, and its <image of a linear map> is the same line:
$$
\ker T=\operatorname{im}T=\operatorname{span}\{(1,0)\}.
$$
Their <set intersection> is nonzero and their sum is only one-dimensional. In particular, $(0,1)$ cannot be written as a sum of one <vector> from each of these <vector subspaces>. The nonzero <set intersection> also prevents an internal <direct sum of vector spaces>. Thus
$$
\boxed{\mathbb R^2\ne\ker T\oplus\operatorname{im}T.}
$$
The <matrix> is a nonzero <nilpotent matrix>; its failure of <diagonalizability> explains why the preceding result does not apply.
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