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.

  • 1
    $\begingroup$ Please show your own efforts. $\endgroup$
    – user0102
    Jun 29, 2020 at 3:01
  • $\begingroup$ 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. $\endgroup$
    – user803264
    Jun 29, 2020 at 3:06
  • $\begingroup$ @JohnP. sorry for confusion. Here, $\phi$ is not zero. $\endgroup$
    – Onyu Kim
    Jun 29, 2020 at 3:44
  • $\begingroup$ Yes. Consider $T:i\mapsto j\mapsto 0$. $\endgroup$ Jun 29, 2020 at 4:42

1 Answer 1


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 $\dim V\ge2$.

  • $\begingroup$ Thank you for your answer. Does your answer hold for a linear transform with simple eigenvalue of zero? $\endgroup$
    – Onyu Kim
    Jun 29, 2020 at 5:59
  • $\begingroup$ @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. $\endgroup$
    – user1551
    Jun 29, 2020 at 6:25
  • $\begingroup$ Thanks. I do not fully understand your comment. Could you give me more explanations about the minimal polynomial of $𝑇$? $\endgroup$
    – Onyu Kim
    Jun 29, 2020 at 6:57
  • $\begingroup$ I know what a minimal polynomial is. I want to know why $x^2$ must divide the minimal polynomial. $\endgroup$
    – Onyu Kim
    Jun 29, 2020 at 7:45
  • $\begingroup$ @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 $m$. $\endgroup$
    – user1551
    Jun 29, 2020 at 9:15

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