Let $V$ be a finite vector space, and let $T$ be a linear transformation $T:V\rightarrow V$. If $\operatorname{null}(T)=\operatorname{span}\{\phi\}$, can $\operatorname{ran}(T)$ contains $\phi$, where $\phi$ is not the trivial vector?
I know that
- $\operatorname{ran}(T)^0=\operatorname{null}(T^*)$ and
- $\operatorname{null}(T)^0=\operatorname{ran}(T^*)$,
where $T^*$ is the dual operator $T^*:V^*\rightarrow V^*$.
Let $\{\phi, e_1, e_2\}$ be a basis in $V$. Then, $\{T(e_1), T(e_2)\}$ spans $\operatorname{ran}(T)$ and there are unique numbers $a_i,b_i$ such that $T(e_1)=a_0\phi+a_1e_1+a_2e_2$ and $T(e_2)=b_0\phi+b_1e_1+b_2e_2$, because $\operatorname{ran}(T)\subset V$.
Now let $\operatorname{null}(T^*)=\operatorname{span}\{\phi^*\}$ then $\phi^*(T(e_1))=\phi^*(T(e_2))=0$. If $\phi^*$ is one element of dual basis such that $\phi^*(\phi)=1$, then $a_0$ and $b_0$ must be zero, and the range does not contain the null space. Moreover $V=\operatorname{null}(T)\oplus\operatorname{ran}(T)$. However I do not know that $\phi^*(\phi)=1$ always.
I have been stuck here.
Best Answer
In general, if $\ker(T)\subseteq TV$, then $\operatorname{rank}(T)\ge\operatorname{nullity}(T)$ and hence $\dim V\ge2\operatorname{nullity}(T)$.
Conversely, if $K$ is any subspace of $V$ such that $n=\dim V\ge2\dim K=2k$, then $r:=n-k\ge k$. Let $\{u_1,u_2,\ldots,u_k\}$ be any basis of $K$. Complete it to a basis $\{u_1,u_2,\ldots,u_k,v_1,v_2,\ldots,v_r\}$ of $V$. Since $r\ge k$, we may define a linear transformation $T$ such that \begin{cases} T(u_i)=0,\\ T(v_i)=u_i&\text{ when }i\le k,\\ T(v_i)=v_i&\text{ when }i> k. \end{cases} Now $K=\ker(T)\subseteq TV$.
In your case, since $K=\operatorname{span}(\phi)$ is one-dimensional, there exists a linear transformation $T:V\to V$ such that $K=\ker(T)\subseteq TV$ if and only if $\dim V\ge2$.