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Let $T:\ V\rightarrow W$ be a linear transformation, if $T$ is bijective and $\{v_1,...,v_n\}$is a basis for $V$, how to show $\{T(v_1),...,T(v_n)\}$ is a basis for $W$?

Here is my thinking process:

$\Rightarrow$

Since $T$ is a linear transformation, $T(a_1v_1+...+a_nv_n)=0\implies a_1T(v_1)+...+a_nT(v_n)=0 $. Since $\{v_1,...,v_n\}$is a basis for $V$, it's linearly independent. Then $a_1v_1+...+a_nv_n=0\implies a_1=...=a_n=0$. Therefore $\{T(v_1),...,T(v_n)\}$ is linearly independent. Since $\dim V=\dim W=n$, it's also a basis for $W$

$\Leftarrow$

If $\{T(v_1),...,T(v_n)\}$ is a basis and $T$ is a linear transformation, then $a_1T(v_1)+...+a_nT(v_n)=0\implies T(a_1v_1+...+a_nv_n)=0$. Since it's also bijective, so injective, so $a_1v_1+...+a_nv_n=0$. Then $a_1=...=a_n=0$ because $\{T(v_1),...,T(v_n)\}$ is linearly independent. Also $\dim V=\dim W=n$, so $\{v_1,...,v_n\}$is a basis for $V$

I feel the order is incorrect. Could someone fix it?

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  • $\begingroup$ What do you show two directions if your question mentions only one? $\endgroup$ – DonAntonio Mar 5 '16 at 22:31
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You want to show that $\{T(v_1),\dots,T(v_n)\}$ is a basis for $W$. There is no double implication to show, but you rather have to prove that

  1. $\{T(v_1),\dots,T(v_n)\}$ is linearly independent
  2. $\{T(v_1),\dots,T(v_n)\}$ is a spanning set for $W$

using the hypothesis that $T$ is linear and bijective.

Proof for 1. Suppose $a_1T(v_1)+\dots+a_nT(v_n)=0$. Then, by linearity, $$ T(a_1v_1+\dots+a_nv_n)=0 $$ Since $T(0)=0$, injectivity forces $a_1v_1+\dots+a_nv_n=0$ and so $a_1=\dots=a_n=0$.

Proof for 2. Let $w\in W$; then $w=T(v)$, for some $v\in V$, by surjectivity. Since $v=a_1v_1+\dots+a_nv_n$, you can conclude by applying $T$.

Fill in the details.


However, also the converse is true: if $\{T(v_1),\dots,T(v_n)\}$ is a basis of $W$, then $T$ is bijective. This follows easily from the rank nullity theorem.

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