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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.

Community
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user450820
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    Which one do you think is correct, and why? – projectilemotion May 31 '17 at 08:47
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    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 ans is option d..I am not able to understand by option c is incorrect. – user450820 May 31 '17 at 08:50
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    C does not imply that T is bijective. It is true for any general T. – Landon Carter May 31 '17 at 08:56
  • There is an error in the question, because both (b) and (d) are correct. $T$ is bijective if and only if it is surjective, which is true when its rank is $n$. See Solomonoff's answer below. – user49640 May 31 '17 at 17:20

5 Answers5

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$(d)$ \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.

Bijesh K.S
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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.

Reinstate Monica
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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.

florence
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  • Your first sentence is a very good insight. Reading the original question and questioner's comments, it's clear they are assuming $T$ is a bijection and trying to prove a)–d). That's the opposite of the question. – Matthew Leingang May 31 '17 at 13:59
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Option c) is incorrect because the null map is not bijective, but assertion c) holds for that map.

José Carlos Santos
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The statement in part (c) is always true; this is the rank-nullity theorem!

lokodiz
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