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So the question asks: Let V be a finite-dimensional vector space, and let T : V → V be a linear transformation. Prove that if T is surjective, then T is an isomorphism. Show that the statement may be false if V is not finite-dimensional.

So so far I got:

Since T is surjective, ∀v1 ∈ V, ∃v2 ∈ V : T(v2) = v1.

T carries linearly independent subsets of V1 onto linearly independent subsets of V2.

Assume that T(x) = 0.

If the set {x} is linearly independent, then by assumption {0} is linearly independent, which is a contradiction.

Hence, the set {x} is linearly dependent.

Then x = 0. That is, N(T) = {0}. Therefore, T is one-to-one.

Since T is onto and one-to-one,

So T is isomorphism.

But I think the proof will still hold if they are not finite -dimensional.Why is T not isomorphism in that case?

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  • $\begingroup$ For surjective part, it should be $T(v2) = v1$. $\endgroup$ – crbah Mar 8 '16 at 11:18
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Use the dimensions theorem: if $\;\dim V=n\;$ and $\;T:V\to V\;$ is a linear map, then

$$n=\dim\ker T+\dim\text{ Im }T$$

Thus, we get that

$$T\;\;\text{is surjective}\;\iff\dim\text{ Im }T=n\stackrel{\text{ by above}}\iff\dim\ker T=0$$

and thus $\;T\;$ is surjective iff it is injective iff it is bijective.

Now, let $\;V=\{\;\{x_n\}_{n=1}^\infty\subset\Bbb R\}\;$ be the real linear space of all real sequences, with sum and scalar multiplication indexwise, and define

$$T:V\to V\;\;,\;\;\;\;T\{x_n\}:=\{x_2,x_3,....\}$$

Clearly $\;T\;$ is surjective but it is not injective.

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