Let $T\colon V\to V$ be a linear transformation such that $\dim(V)=n<\infty$. Prove that

(a)$T$ is bijection


(b)T is injective.


  1. show $(a)\implies(b)$

If $T$ is bijection, then it is injective by definition.

  1. show $(b)\implies (a)$

Let $T$ be injective, then if $v_1\neq v_2 \implies T(v_1) \neq T(v_2)$. Hence, $\exists w\in V$ such that $T(v) \neq w$ for all $v \in V$

Therefore, $\dim(\operatorname{Image}(T))<n \implies \dim(\ker(T))>1$.

So, $\exists z\in V$ such that $T(z)=0$.

Moreover, Let $v$ be any nonzero vector in $V$, $T(v+z) = T(v)+T(z) = T(v)$.

$T$ is not injective, a contradiction.

Is my solution correct?

  • $\begingroup$ I don't think this argument is correct. When you assume that T is injective, you can't conclude that there is a w in V such that $T(v)\ne w$ for all $v\in V$. (This would mean that T is not surjective.) $\endgroup$ – user84413 Apr 27 '15 at 16:34

I think it is essentially correct, but I would write things a bit differently. You are right on using rank-nullity. But then I would say that the only vector $z$ with $T(z)=0$ is the zero vector. Then the kernel is 0-dimensional, i.e., nullity is $0$, so rank is $n$, and then $T(V)$ is an n-dimensional subspace of $V$, and so it must be the whole of $V$ itself (n-dimensional vector spaces do not have proper n-dimensional subspaces.) Then $T(V)=V$. Since the kernel is $0$ , the map is injective , as you said, f T(z)=T(w); $z \neq w$, then $T(z-w)=0$. Then $T(V)=V$, i.e., $T$ is onto and it is injective.

Just a few things tidied up a bit, but overall correct, I would say.

  • $\begingroup$ I followed your approach until "Since the kernel is 0...". Can you please clarify? @gary $\endgroup$ – user3382078 Apr 27 '15 at 14:55
  • $\begingroup$ Sure, @user234784 : say the kernel is $0$. Then $T$ must be injective; if it was not, we would have $z \neq w $ with $T(z)=T(w)$ But then by linearity,$T(z-w)=0$, contradicting the assumption that the kernel is $0$.In the other directio, if $T$ is not injective, then there are $z,w$ with $ T(z)=T(w)$ so that, by linearity, $T(z)-T(w)=0=T(z-w)$, i.e., the kernel is not trivial if $T$ is not injective. $\endgroup$ – gary Apr 27 '15 at 15:02

The formula $\dim(\operatorname{Ker} T)+\dim(\operatorname{Im} T)=\dim V=n$ from the "rank + nullity theorem" should be enough for this.

If $T$ is injective, then $\dim( \operatorname{Ker} T) =0$ so $\dim(\operatorname{Im} T) =n$, therefore since $\operatorname{Im} T$ is a subspace of $V$ you get $\operatorname{Im} T=V$, because the only subspace of $V$ with the same dimension is $V$ itself.

The inverse is easier: a bijection is always, by definition, injective.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.