Rank nullity theorem -bijection 
Let $T:\Bbb R^n\to \Bbb R^n$ be a linear transformation. Which of the following statement implies that $T$ is bijective?
a) $\operatorname{Null}(T)=n$
b) $\operatorname{Rank}(T)=\operatorname{Null}(T)=n$
c) $\operatorname{Rank}(T)+\operatorname{Null}(T)=n$
d) $\operatorname{Rank}(T)-\operatorname{Null}(T)=n$

I think that since $T$ is one one n onto... Nullity will be zero... So option a) and b) are incorrect..
And c) and d) will be correct.. But the answer is option d). I am not able to understand why option c) is incorrect.
 A: $(d)$
\begin{align}
R(T)+N(T)=n\tag1\\
R(T)-N(T)=n\tag2
\end{align}
We know $(1)$ holds anyway due to rank-nullity theorem. Thus solving $(1)$ and $(2)$ we get $R(T)=n$ and $N(T)=0$ which implies $T$ is a bijection.
$(c)$ is an incorrect choice. Consider $T:\mathbb{R}^2\to \mathbb{R}^2$ such that $T(x_1,x_2)=(0,x_1).$ $R(T)+N(T)=n$ will anyway hold. But this $T$ is not a bijection.
A: In addition to option (d), option (b) is correct. If $n = 0$ then the consequence, $T$ is bijective, is true. On the other hand if $n \ne 0$ then the premise, that the rank equals the nullity, is false. Either way the implication is true.
A: The question isn't asking you which of the options will be the case if $T$ is a bijection, it's asking which of those prove that $T$ is a bijection. Your reasoning for the first two being incorrect is valid. $c$ certainly doesn't imply that $T$ is a bijection, since it is just the statement of the rank-nullity theorem (for maps from a finite dimensional vector space to itself), which holds for non-invertible transformations. $d$ does imply that $T$ is a bijection, since if the nullity were greater than zero then the rank would be greater than $n$, which is impossible; thus, $T$ is injective, and therefore surjective. 
A: Option c) is incorrect because the null map is not bijective, but assertion c) holds for that map.
A: The statement in part (c) is always true; this is the rank-nullity theorem!
