# Linear Algebra - Proof on Linear Transformations and Independence

Let $$T: \Bbb{R}^n \rightarrow \Bbb{R}^n$$ be an invertible linear transformation. Prove that if {$${ \vec{v_1},\vec{v_2},...,\vec{v_n}}$$} is a linearly independent set of $$\Bbb{R}^n$$ if and only if {$$T( \vec{v}_1),T( \vec{v_2}),...,T( \vec{v_k})$$} is a linearly independent set of $$\Bbb{R}^n$$.

So I know that for a set of vectors to be linearly independent then they cannot be written as a linear combination of one another. So can I say something along the lines of $$t_1 v_1 + t_2 v_2 + ... + t_k v_k = 0$$ then taking the transformations the set I get is {$$T( t_1 \vec{v}_1),T( t_2 \vec{v_2}),...,T( t_k \vec{v_k})$$} which given the definition of linear independence can be written as $$t_1 T( \vec{v}_1) + t_2 T( \vec{v_2}) +,..., + t_k T( \vec{v_k}) = 0$$. So then is that it?

• Yes. You have the right line of thinking. Be careful:you've only shown that a linear combination in the domain implies a linear combination in the range. You may want to remark why none of the images T(v$_i$) can be 0 in this situation. Otherwise, a non-trivial solution may map to a trivial solution if certain images are 0. – Joel Pereira Oct 22 '18 at 4:14

For $$\Longrightarrow$$ : You are on the right way!
$$t_1T(v_1)+t_2T(v_2)+\cdots+t_nT(v_n)=0$$ implies $$T(t_1v_1+\cdots+t_nv_n)=0=T(0)$$ and so by one-one of $$T$$, $$t_1v_1+\cdots+t_nv_n=0$$
and by independence of $$v_i$$, $$t_1=t_2=\cdots=t_n=0$$
For $$\Longleftarrow$$ : Let $$\alpha_1v_1+\alpha_2v_2+\cdots+\alpha_nv_n=0$$ and apply $$T$$ on both sides and use independence of $$T(v_i)$$'s to get the result!