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Let $T:\ V\rightarrow W$ be a linear transformation. Suppose $\dim V=\dim W$ and $\{v_1,\ldots,v_n\}$, $\{w_1,\ldots,w_n\}$ are bases for $V$ and $W$ respectively. Let $T$ be defined by $T(v_i)=w_i,\ i=1,\ldots,n$. How to show $T$ is bijective?

My try:

Since $\dim V=\dim W$, we only need to show either injectivity or surjectivity. Since $T(v_i)=w_i$, $\forall y\in W, \exists x \in V$ such that $T(x)=y$. Therefore we have surjectivity. Therefore $T$ is bijective.

I think I missed something but wouldn't know what. Could someone help fix the proof?

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Your proof is "correct" but skips over a step that in an introductory course you might want to make explicit.

Namely, if $y \in W$, then $y = \sum_{i=1}^n a_i w_i$ for some scalars $a_i$. Now set $x = \sum_{i=1}^n a_i v_i$. Then $T(x) = \sum_{i=1}^n a_i T(v_i)$, because $T$ is linear (this is the crucial insight). Therefore $T(x) = y$, which completes the proof that $T$ is surjective.

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