# Thread: show that the set is linerlly independent

1. ## show that the set is linerlly independent

Suppose that ${\mathbf{v}_1,\mathbf{v}_2...\mathbf{v}_k}$ is linearly independent oset of vectors in $\mathbb{R}^n$. Show that if A is an nxn nonsingular matrix then ${A\mathbf{v}_1,A\mathbf{v}_2...A\mathbf{v}_k}$ is also linearly independent.

So if the given set is linearlly independent that means that $c_1 \mathbf{v}_1+c_2 \mathbf{v}_2+...+c_k \mathbf{v}_k=\mathbf{0}$
and that $c_1=c_2=...=c_k=0$
doesn't that mean that A must be 0?

can I use the theorem that if the larger set is linearally independent then it's subspace must be linearally independent? I'm not sure if I can use this becuse it's not like Av is part of the original set of vectors.

2. I'm not sure what you're allowed to use at this point, but do you have any results about a linear map and the dimension of its domain, image, and kernel?

3. Originally Posted by superdude
Suppose that ${\mathbf{v}_1,\mathbf{v}_2...\mathbf{v}_k}$ is linearly independent oset of vectors in $\mathbb{R}^n$. Show that if A is an nxn nonsingular matrix then ${A\mathbf{v}_1,A\mathbf{v}_2...A\mathbf{v}_k}$ is also linearly independent.

So if the given set is linearlly independent that means that $c_1 \mathbf{v}_1+c_2 \mathbf{v}_2+...+c_k \mathbf{v}_k=\mathbf{0}$
and that $c_1=c_2=...=c_k=0$
doesn't that mean that A must be 0?

can I use the theorem that if the larger set is linearally independent then it's subspace must be linearally independent? I'm not sure if I can use this becuse it's not like Av is part of the original set of vectors.
If a set $\{v_1,...,v_k\}$ is linearly independent then $c_1v_1 + c_2v_2 + ... + c_kv_k = 0 \Leftrightarrow c_1 = c_2 = ... = c_k = 0$ where $c_1, c_2,...,c_k$ are scalars.

Now, note that for any scalars $a_1,a_2,...,a_k$, we have that $a_1Av_1 + a_2Av_2 + ... + a_kAv_k = Aa_1v_1 + Aa_2v_2 + ... + Aa_kv_k = A(a_1v_1 + a_2v_2 + ... + a_kv_k)$

Now use the fact that A is nonsingular to finish.