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Let $V$ be a finite dimensional vector space. Show that a linear transformation $T\colon V \to V$ is one-to-one if and only if it is onto.

The hint I was given was that one only needs to show that $T(e_1),\dots,T(e_n)$ is a basis whenever $e_1,\dots,e_n$ is a basis.

My first attempt is as follows: Let $x \in V$. Then $$Tx=[T(x_1)e_1+\dots+T(x_n)e_n]=[x_1T(e_1),\dots,x_nT(e_n)]=x\cdot T(e).$$ I really want $T(e_n)$ to be a basis of $V$, but I'm not sure that it is. It seems like it is obviously a basis because $T$ maps from $V$ to $V$.

Is there a property of linear transformations that makes this apparent?

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$T$ maps from $V$ to $V$, but not necessarily to all of $V$. If transformation $T$ is a singular matrix, then it will map to a subspace of $V$ with fewer dimensions. Thus multiple vectors in $V$ will collide to the same transformed vector. This is "projection" (because a 3D object being flattened to a 2D view is a case of this). – Kaz Apr 8 '13 at 6:58
If you show that the transformation is one-to-one iff the transformation matrix is invertible, and if you show that the transformation is onto iff the matrix is invertible, then by transitivity of iff you also have iff between the one-to-one and onto conditions. – Kaz Apr 8 '13 at 7:00
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How do you normally check for a basis? Well, try that! A good definition for this problem is: For every $v \in V$ we can write $v$ as a linear combination of $e$ (your basis).

$T$ is one-to-one: $T[e_i] \neq T[e_j]$, use that to check for the basis.

$T$ is surjective: $v \in V$, then there is an $w \in V$ such that $T[w]=v$. Now write $w$ in its basis $e$ and use that $T$ is linear.

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Well if $e_1,...,e_n$ is a basis then $c_1e_1+...+c_ne_n=0$ iff $c_i$ for all $i$ is $0$ since each $e_i$ is linearly independent. Since $T$ is a linear transformation $T(ce)=[T(c_1e_1,...,c_ne_n)=[c_1T(e_1)+...+c_nT(e_n)]$. I'm not sure how to show that is linearly independent and hence a basis. – emka Apr 8 '13 at 6:55
If your using that T is one to one: What is T[0]? Set T(ce) equal to T(0). If you want a more explaining answer, and not hints, you can say so. – Ryan Apr 8 '13 at 6:58
Well since it is a linear transformation $T(0)=0$. – emka Apr 8 '13 at 7:00
Good, now when is T(e)=0? – Ryan Apr 8 '13 at 7:01
$T(e)=0$ whenever $e=0$. – emka Apr 8 '13 at 7:05

The finite-dimension hypothesis is very important, since this result fails otherwise. Thus we must make use of it in our proof.

I'll try and give a rough sketch of the proof, leaving you to fill in the details. Suppose that $T$ is onto. We want to show it is one-to-one. That is, we want to show that if $T(v) = T(w)$, we have $v = w$. Using linearity, we can rewrite this as $T(v-w) = 0$ implying $v -w = 0$, so that the kernel of $T$ is only zero. How does a non-zero kernel contradict onto-ness? Let $u$ be nonzero, but so that $T(u) = 0$. Then we can extend $u$ to a basis for $V$, and the image of this basis must still form a spanning set, since $T$ is onto. But the image of this basis contains a zero, so we can find a smaller spanning set, impossible in finite dimensions.

Suppose now that $T$ is one-to-one. This means, as before, that if $T(v) = 0$ then $v = 0$. How can we prove that $T$ is onto? Extend $v$ to a basis. The image of this basis by $T$ must still be linearly independent (since $T$ was one-to-one). But if it does not span, we can add a vector to it not in its span and it will still be linearly independent. Now we have a linearly independent set larger than the dimension of the space, impossible in finite dimensions.

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Yes $\{T(e_1),T({e_2}),.....,T(e_n)\}$ will be a basis for V because :

if $x_{1}T(e_1)+.....+x_{n}T(e_n)=0$ then $T(x)=0$ such that $x=(x_1,....,x_n)$


and it is a theorem at linear algebra that if T is one to one Then $\operatorname{ker} n(T)=\{0\}$ and so $x=0$ also $x_1=......=x_n=0$

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Hint: Let us write $W:=V$ to make this less confusing. Then, you have an injective linear transformation $T:V\to W$. To prove that it is necessarily surjective, use the fact that injectivity implies $\dim T(V)=\dim V=\dim W$.

To show that surjectivity implies injectivity, note that if $\ker T$ was non-trivial, then $\dim \text{im}(T)$ would necessarily be less than $\dim V$. Since $\dim V=\dim W$, to see a problem with this?

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Alex's answer is very good, but matrices are nice too. Represent $T$ by $M$, a square matrix. Suppose that $\sum \lambda_i Me_i = 0$. If $M$ is 1-to-1 then it's invertible. For onto $\Rightarrow$ injective, consider the rank.

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