If a Vector Space $V$ has dimension $n$ and $T:V\to V$ is linear, then is the following true?

Question

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?

Answer 1

Suppose $T(v_1) = v_1, \quad T(v_2) = v_2, \quad T(v_3) = v_1+v_2.$ Then $\{v_1,v_2\}$ is a basis of $\operatorname{im}(T)$ but $\{v_3\}$ is not a basis of $\ker(T).$ In this case $\{v_1+v_2-v_3\}$ is a basis of $\ker(T).$

Answer 2

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 $\{0\}$.

Answer 3

Why should it, even if $T$ is idempotent? As a counter-example, consider the subspace $U$ generated by $v_1,\dots, v_k$ and let $T$be the projection from $V$ onto $U$. Its kernel is indeed generated by $v_{k+1},\dots,v_n$, BUT $\;(v_{k+1}+v_1, \dots ,v_{n}+v_1)\;$ is a basis of another complement of $U$, and it is not contained in $\ker T$.