Is $V \oplus \{0\}$ a direct sum? Consider this statement:
If a linear operator $N:V \to V$ is nilpotent, $N^k=0$, and $U$ is a subspace with $U \cap\ker(N^{k-1})=\{0\}$ then there is a subspace $W$ with $NW \subseteq V$ and $$ V = W \oplus (U + NU + N^2 U + ... + N^{k-1}U)$$ 
It is proved by induction on dimension of $V$. Consider the case $\dim(V)=1$. The only subspaces of $V$ are $\{0\}$ and $V$. No? Because we have $U \cap\ker(N^{k-1})=\{0\}$, in this case $U=\{0\}$ and $W=V$. 
But if $V \oplus \{0\}$ is a direct sum we can always choose $U=\{0\}$ in the statement and it is trivial. Should the statement say "proper subspace $U$ and $W$"? But then the case in dimension one is false. Please help me understand. 
 A: Any sum of subspaces of the form $X+\{0\}$ is a direct sum, as follows immediately from the definition of that notion. However in the given statement, $U$ is introduced in the hypothesis, not in the conclusion, so it is given to you, and one does not have the liberty to replace it by$~\{0\}$. Since the case $U=\{0\}$ is so easy, the statement might have said "$U$ is a nonzero subspace". But it is in general a bad idea to put effort into excluding cases just because they are uninteresting, when the statement formulated actually remains valid in that trivial case; the explicit exclusion would suggest that it is necessary, in other words that the given statement would fail (uninterestingly) in the excluded case. In the current example, adding "nonzero" would suggest that the proof somewhere needs to use that $U\neq\{0\}$, while in fact there is no such need here.
As examples where, contrary to the situation here, the explicit exclusion of trivial cases is called for (I've emphasised the exclusion): "every non-constant complex polynomial has at least one root", and related to this, "any linear operator on a finite non-zero dimensional complex vector space has at least one eigenvalue".
