# Can the range of a linear transformation contains the null space?

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.

• Yep! For example, let $T$ be the one and only linear operator on the trivial space. Then we can take $\phi = 0$, and $\operatorname{null}(T) = \{0\} = \operatorname{span}(\phi)$, but $\phi = 0 \in \operatorname{ran}(T)$ as well. Jun 29, 2020 at 3:06
• @JohnP. sorry for confusion. Here, $\phi$ is not zero. Jun 29, 2020 at 3:44
• Yes. Consider $T:i\mapsto j\mapsto 0$. Jun 29, 2020 at 4:42

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 $$\dim V\ge2$$.
• @OnyuKim If zero is a simple eigenvalue and $\ker(T)=\operatorname{span}(v)$, the range of $T$ cannot contain $\operatorname{span}(v)$, for, if $Tu=v$, then $x^2$ must divide the minimal polynomial of $T$ and hence $0$ is not a simple eigenvalue. Jun 29, 2020 at 6:25
• Thanks. I do not fully understand your comment. Could you give me more explanations about the minimal polynomial of $𝑇$? Jun 29, 2020 at 6:57
• I know what a minimal polynomial is. I want to know why $x^2$ must divide the minimal polynomial. Jun 29, 2020 at 7:45
• @OnyuKim Note that $u$ and $v$ must be linearly independent, because $Tu=v\ne0=Tv$. Therefore, the monic polynomial $p$ of the least degree such that $p(T)u=0$ is $p(x)=x^2$. However, the minimal polynomial $m(x)$ of $T$ also satisfies the condition that $m(T)u=0$. If $p$ doesn't divide $m$, then by Euclidean algorithm, $a(x)p(x)+b(x)m(x)=c(x)$, where $c\ne0$ is the gcd of $p$ and $m$ and $a,b$ are some polynomials. It follows that $c(T)u=0$. But this contradicts the definition of $p$, because $\deg c<\deg p$. Hence $p$ must divide $m$. Jun 29, 2020 at 9:15