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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.

My attempt: By definition, surjection is that every element on the codomain is covered. Therefore, if every element on T is covered, and since it is mapping to itself, it must be an isomorphism (also by definition).

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Maybe rank-nullity theorem. If $T$ is surjective it has full rank, which means (by the rank-nullity theorem) that it has a trivial null-space so it is injective. – User8128 Mar 11 at 4:12
    
Use the finite basis and proceed as you would with a surjective map between finite sets of the same size as well as the properties of linearity. – qbert Mar 11 at 7:24

Establish the following facts:

  1. (Rank-Nullity Theorem) $\dim \ker T + \dim \operatorname{im} T = \dim V$. For this you need to explicitly define a basis.

  2. Since $T$ is surjective, $\dim \operatorname{im} T = \dim V$, so $\dim\ker T = 0$

  3. If $\dim \ker T = 0$, then $T$ is injective. (Show that $\dim\ker T = 0 \iff \ker T = \{0\}$)

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By Rank-Nullity theorem $$\dim\mathbb{V}=\dim ker(T)+\dim\mathbb{V}$$ so $$dim ker(T)=0.$$ Then $T$ is injective and therefore $T$ is a isomorphism.

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You are wrong that a (linear) mapping from a space to itself that covers every element (i.e., which is surjective) is an isomorphism by definition. By definition an isomorphism must admit a morphism that is inverse to it, and this is only possible if the map is (also) injective. So what you must prove is that in the given setting a surjective linear map is automatically injective too. (After proving this you have a linear map that is bijective, therefore has an inverse map; it is a standard exercise that such an inverse map is automatically linear.) In proving this you must use that $V$ is finite dimensional, since the claim is false in infinite dimensional vector spaces (for instance in the space of polynomial functions on$~\Bbb R$, taking the derivative is a surjective linear transformation that is not injective: functions that differ by a constant have the same image).

This is easy once you now that finite dimension is well defined, namely that is a vector space $V$ has some finite basis then every basis of $V$ has the same number of elements; proving this requires work, but is standard theory that I am not going to repeat here. Assuming this, let $[e_1,\ldots,e_n]$ be some basis of $V$, and chose $b_1,\ldots,b_n$ such that $T(b_i)=e_i$; you can do this because $T$ is surjective. Now $b_1,\ldots,b_n$ are linearly independent, since a relation $c_1b_1+\cdots+c_nb_n=0$ for scalars $c_1,\ldots,c_n$ implies (by applying $T$ to both sides, and using linearity) that $c_1T(b_1)+\cdots+c_nT(b_n)=0$, in other words $c_1e_1+\cdots+c_ne_n=0$, and this can only happen when all $c_i$ are zero. One can extend any linearly independent family to a basis, so we can add some number $k$ of vectors to $[b_1,\ldots,b_n]$ to obtain a basis of $V$. But then we have a basis of $n+k$ elements, and since we also have a basis $[e_1,\ldots,e_n]$ of $n$ elements, we conclude $k=0$ by the fact that dimension is well defined; this means that $[b_1,\ldots,b_n]$ was already a basis of$~V$.

But then every vector of $V$ is a linear combination of $[b_1,\ldots,b_n]$, and $T$ is injective because we already showed that the only such linear combination mapped to$~0$ by$~T$ is the zero combination. In fact we don't even need to mention injectivity separately here, because we can explicitly give the the inverse linear transformation $S$ to$~T$, which is defined on the basis $[e_1,\ldots,e_n]$ by $S(e_i)=b_i$ (and extended by linearity). By construction $T\circ S$ maps $e_i\mapsto e_i$ and so is the identity on $V$ (this does not use that $[b_1,\ldots,b_n]$ is a basis); the other composition $S\circ T$ maps $b_i\mapsto b_i$ and so, as $[b_1,\ldots,b_n]$ is a basis, is the identity on $V$ too.

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Suppose $T\colon V\to V$ is not injective. Let $\{v_1,\dots,v_m\}$ be a basis of $\ker T$ and complete it to a basis $\{v_1,\dots,v_m,v_{m+1},\dots,v_n\}$ of $V$.

Then, by assumption, $m>0$ and $\{T(v_{m+1}),\dots,T(v_n)\}$ is a spanning set for the image $\operatorname{im}T$ of $T$. Therefore $\dim\operatorname{im}T\le n-m<n$ and $T$ is not surjective.


Note that this requires finite dimensionality of $V$ and, indeed, the statement is not true for infinite dimensional vector spaces. So your argument is incorrect, otherwise it would apply also in the infinite dimension case.

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