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Let $L: V \rightarrow W$ be an injective linear transformation.

Let $dim V , dimW = n$.

Show that L is surjective.


My thoughts:

If $L(V)$ is the image of $V$, we can show that that $L: V \rightarrow L(V)$ is a bijection and thus that $dim L(V) = n$. And then from there maybe you can show that L(V) has to be equal to W (surjective) maybe with a proof by contradiction but I can't quite make it work...

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You have the right idea. Any $n$-dimensional subspace of an $n$-dimensional vector space must be the whole space. Why? Well, assume you have $w \in W$ that is not in $L(V)$. Then you can add $w$ to a basis of $L(v)$ to form a new subset of $W$. However, this subset now has $n+1$ linearly independent vectors, a contradiction.

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Hint:

Take a basis $v_1,\dots,v_n$ of $V$, and show that $L(v_1),\dots,L(v_n)$ is a basis for $W$.

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