# How to show $\dim \operatorname{range}T=\dim\operatorname{null}R$?

Is $\operatorname{range}T$ a subspace of $\operatorname{null}R$?

The above is the question I am referring to.

Sorry I have some problem with proving $\dim \operatorname{range}T=\dim\operatorname{null}R$.

I now have \begin{align}\dim\operatorname{range}T & =\dim \operatorname{range}T'=\dim P_5(\mathbb R)-\dim\operatorname {null}T'\\ & = \dim P_5(\Bbb R)-\dim\operatorname{span}(R) = 6-1=5.\end{align}

However, we know that $\dim\operatorname{null}R=\dim P_5(\Bbb R)-\dim \operatorname{range}R$. In this case, $\dim\operatorname{range}R$ does not have to be $1$. It may be zero, in which case $R$ is the zero map. Therefore I am confused about how to show $\dim\operatorname{range}T=\dim\operatorname{null}R$.

This question is related to the following general statement: Suppose $V$ and $W$ are finite-dimensional, $T$ is a linear transformation from $V$ to $W$, and there exists $R$ which is a linear functional on $W$ such that $\operatorname{null}T'=\operatorname{span}(R)$. Then $\operatorname{range}T=\operatorname{null}R$.

You are right to say that $\operatorname{range}R$ can have dimension $0$ or $1$; but you have forget that the same is true for $\operatorname{span}(R)$. Indeed, if $R=0$ then $\operatorname{span}(R)=\{0\}$. In particular, $$\dim \operatorname{span}(R)=\dim \operatorname{range} R,$$ which is exaclty what you need to finish your proof.