# Contradicting the non-existence of a linear map $T: \Bbb R^5 \to \Bbb R^5$ and the Fundamental Theorem of Linear Algebra (from Axler Exercise 3.B(5))

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 $$T:\Bbb R^5 \to \Bbb R^5$$)

$$e_3\,-\,e_4$$ is also in the null space of your $$T$$, but is not in the span of $$e_1, e_2$$.