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

(a)$T$ is bijection

iff

(b)T is injective.

Solution:

  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?

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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.

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