Show $\{ T(v_1),\ldots,T(v_n) \}$ is linearly independent Let $\{v_1,\ldots,v_n\}$ be a linearly independant set of vectors spanning $\mathbb{R}^n$, and let $T \in L(\mathbb{R}^n,\mathbb{R}^n)$ where $T$ is also invertible.
Prove that  $\{ T(v_1),\ldots,T(v_n) \}$ is linearly independent if and only if $T$ is invertible.
I dont really know how to approach this.
I figure i can prove one direction by setting $T$ to be invertible then maybe using the defintions of one-to-one and onto...but i am still not sure how to apply this.
 A: Consider a dependence $c_1T(v_1)+\dots+c_nT(v_n)=0$.  By linearity, this gives $T(c_1v_1+\dots+c_nv_n)=0$.
On one hand, if $T$ is invertible, then it is one-to-one, so $c_1v_1+\dots+c_nv_n=0$.  By independence of these vectors, we have $c_1=\dots=c_n=0$, and so $\{T(v_1),\dots,T_(v_n)\}$ is independent.
On the other... $T$ need not be invertible if $\{v_1,\dots,v_n\}$ is merely an independent set and not also a spanning set.  For instance, $T:\mathbb{R}^2\to\mathbb{R}^2$ defined by $T(x,y)=(x,0)$ with the independent set $\{(1,0)\}$ -- this has $\{T(1,0)\}$ an independent set, too.  
Though if we do go and assume $\{T(v_1),\dots,T(v_n)\}$ is a spanning set as well, then for any vector $w\in V$ we may write it as $w=c_1 T(v_1)+\dots+c_n T(v_n)$ for some scalars $c_1,\dots,c_n$, which are unique by independence.  By linearity, $w=T(c_1v_1+\dots+c_n v_n)$.  Thus, $T$ is invertible: this $c_1v_1+\dots+c_nv_n$ is $T^{-1}(w)$, and this can be done for each $w\in V$.
A: If there is some set of non trivial $a_1,....,a_n$ with 
$a_1T(v_1)+.....+a_nT(v_n)=0$, then  
$T(a_1v_1)+.....+T(a_nv_n)=0\Rightarrow 
T(a_1v_1+.....+a_nv_n)$ all by the linearity of $T$. Now what can you say about the kernel of an invertible linear map?
A: Suppose $T$ is invertible, and suppose 
$$c_1 T(v_1) + c_2 T(v_2) + \cdots c_n T(v_n) = \vec0.$$
The goal is to prove that all the $c_i$'s are zero. But $T$ is a linear transformation, so  the above equation is equal to
$$T(c_1 v_1 + c_2 v_2 + \cdots + c_n v_n) = \vec 0.$$
$T$ is invertible, so 
$$c_1 v_1 + c_2 v_2 + \cdots + c_n v_n = \vec0.$$
But the $v_i$'s are linearly independent, so all the $c_i$'s are zero. 
For the reverse implication, suppose $T$ is not invertible. That means $T$ is not one-to-one, so there is a nonzero vector $u$ such that $T(u)=\vec0$. Write $$u = c_1 v_1 + c_2 v_2 + \cdots + c_n v_n.$$
Then 
$$\vec0 = T(u) = T(c_1 v_1 + c_2 v_2 + \cdots + c_n v_n)
= c_1 T(v_1) + c_2 T(v_2) + \cdots + c_n T(v_n).$$
Since $u\not=\vec0$, and $\{v_1,v_2,\ldots,v_n\}$ is linearly independent, not all of the $c_i$'s are zero. This implies $\{T(v_1), T(v_2),\ldots, T(v_n)\}$ is linearly dependent.
