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

Here is my try:

Since we know $\{w_1,...w_2\}$ is a basis for $W$ and $T(v_i)=w_i,\ i=1,...n$, $\{T(v_1),...,T(v_n)\}$ is a basis for $W$. $\{T(v_1),...,T(v_n)\}$ is linearly independent and(can't find an imply relationship) $\{v_1,...,v_n\}$ is independent. Then by $T(v_i)=w_i$, $T$ is bijective.

I found my process nonsense. Could someone help?

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So, you showed that $\{T(v_1),\ldots,T(v_n)\}=\{w_1,\ldots,w_n\}$ is a basis for $W$. Therefore, the image of $T$ is the whole $W$, i.e., $\text{dim}(\text{Im}\;T)=n$. Now the result easily follows from the rank-nullity theorem, since $V$ and $W$ have both the same dimension.

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If $T(x) = T(y) \Rightarrow T(x-y) = 0 \Rightarrow x-y \in \text{Ker(T)}= \{0\}$ by the rank-nullity theorm since $\text{dim}(V) = \text{dim}(W)=n < \infty$. Thus $x - y = 0 \Rightarrow x = y$. So $T$ is one-to-one. Its onto because let $w \in W \Rightarrow w = a_1w_1+\cdots a_nw_n = a_1T(v_1)+\cdots a_nT(v_n) = T(a_1v_1+\cdots a_nv_n)$. Thus set $v = a_1v_1+\cdots a_nv_n \Rightarrow w = T(v)$ . So $T$ is onto.

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Suppose

$$x=\sum_{k=1}^n a_kv_k\in\ker T\implies0=Tx=\sum_{k=1}^na_kTv_k=\sum_{k=1}^na_kw_k\implies\;$$

$$a_k=0\;\;\;\forall\,k\implies x=0$$

so that $\;\ker T=\{0\}\iff T\;$ is bijective, since $\;\dim V=\dim W<\infty\;$

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You can proceed as follows:

Surjectivity: any element $y\in W$ can be written (uniquely) as $y=\sum_{i=1}^ny_iw_i$, where $y_i\in K$ (where $K$ is your base field). Then, $y=\sum_{i=1}^ny_iT(v_i)=\sum_{i=1}^nT(y_iv_i)=T(\sum_{i=1}^ny_iv_i)$. Call $x$ the element $\sum_{i=1}^ny_iv_i$ of $V$. Clearly $T(x)=y$. So you have surjectivity.

Injectivity: suppose you have two elements $x,x'\in V$ such that $T(x)=T(x')$. You can write uniquely $x$ and $x'$ as linear combinations of the elements in your basis, so that $$x=\sum_{i=1}^nx_iv_i\ \ \ x'=\sum_{i=1}^nx_i'v_i.$$ Hence, $$T(x)=\sum_{i=1}^nx_iw_i\ \ \ T(x')=\sum_{i=1}^nx_i'w_i.$$ Since $\{w_1,\dots,w_n\}$ is a basis, there is a unique way to write a given element (in our case $T(x)=T(x')$) as linear combination of the $w_i$'s, as a consequence: $x_i=x_i'$ for any $i=1,\dots,n$. But this implies $x=x'$, so that $T$ is injective.

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