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:


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 $W$


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?

  • $\begingroup$ What do you show two directions if your question mentions only one? $\endgroup$ – DonAntonio Mar 5 '16 at 22:31

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.