If a Vector Space $V$ has dimension $n$ and $T:V\to V$ is linear, then is the following true? Suppose we have that a basis for $V$ is 
$B=\{v_1,v_2,\ldots ,v_n\}$ and $v_1,v_2,\ldots,v_k$ are a basis for $\operatorname{im}(T)$.
Does it follow that $v_{k+1}, v_{k+2},\ldots, v_n$ is a basis for $\ker(T)$? What if $T$ is idempotent? 

There is no reason that the first claim be true in general. A counterexample is $n=2$ and $T(v_1)=v_1$ and $T(v_2)=v_1$. Then the image of $T$ is spanned by $v_1$, yet $v_2$ is not in the kernel of $T$.
Even if $T$ is idempotent, this need not be true (actually my above example is idempotent). If $T$ is the projection onto, say, the span of $v_1, \ldots, v_k$, then it is idempotent, but its kernel is $\{0\}$.
