I am asked to prove there does not exist a linear map $T:\Bbb R^5 \to \Bbb R^5$ such that $\operatorname{range}(T) = \operatorname{null}(T)$.
I think I understand the proof whereby the Fundamental Theorem of Linear Algebra (aka rank nullity theorem) shows how this isn't possible. The null space and the range can't each have a dimension of 2.5
But I want to ask what is wrong with the following linear map, why this somehow doesn't work:
Let the basis of $\Bbb R^5$be defined as 5 independent vectors $e_1, e_2, e_3, e_4$and $e_5$.
Define a linear map $T:\Bbb R^5 \to \Bbb R^5$as follows: $Te_1=0, Te_2=0, Te_3=e_1, Te_4=e_1, Te_5=e_2.$
If I define T this way, it seems that $\operatorname{null}(T) = \operatorname{span}(e_1, e_2)$ and $\operatorname{range}(T) = \operatorname{span}(e_1, e_2)$, ie $\operatorname{range}(T) = \operatorname{null}(T)$, which is not supposed to be the case given the textbook is asking me to prove that this isn't possible.
Where I am going wrong here? (For context, the previous problem in this exercise asked to give an example of a showed a similar linear map $T:\Bbb R^4 \to \Bbb R^4$ where $\operatorname{range}(T) = \operatorname{null}(T)$, and supposedly one such linear map $T$ is as as follows: $Te_1=e_3, Te_2=e_4, Te_3=0, Te_4=0$ according to the solutions. Notice that doesn't seem too different to what I'm doing in $T:\Bbb R^5 \to \Bbb R^5$)
Best Answer
$e_3\,-\,e_4$ is also in the null space of your $T$, but is not in the span of $e_1, e_2$.