The image of a basis under an onto linear transformation is a basis I am having a problem with 1 question. Thank you
Suppose $T: \mathbb{R}^n \to \mathbb{R}^n$ is an onto linear transformation.
If $\{v_1,\ldots,v_n\}$ is a basis of $\mathbb{R}^n$, show $\{T(v_1),\ldots,T(v_n)\}$ is also a basis.
Honestly i dont understand this question at all...please help with it, i just dont have any idea how to approach it.
 A: Think it through one step at a time.
What do you have to do? Show something is a basis. How do you do that? First, you show it's linearly independent, then you show it generates the space.
Linear independence. How do you show something is linearly independent? You show that there's no non-trivial linear combination of them that equals zero. How do you show that? Algebra: you write down "linear combination equals zero", and then prove that all the coefficients are zero.
$$\lambda_1T(v_1)+\lambda_2T(v_2)+\dots+\lambda_nT(v_n)=0$$
And you need to show $\lambda_1=\dots=\lambda_n=0$. If you're stuck, look at what the assumptions of the exercise are: that $T$ is linear and onto. Obviously you'll have to use at least one of those assumptions, so that sharply limits what your next step should be.
Now you need to show that it generates the space. This requires a bit more creativity than the first question. Use the fact that in an $n$-dimensional space, any set of $n$ linearly independent vectors generates the space.
A: Well for the second link that you have provided, yo may note the following :
Consider a linear combination of $T(v_1),T(v_2),\ldots, T(v_n)$ such that 
$a_1T(v_1)+a_2T(v_2)+\ldots +a_nT(v_n)=0$
By linearity of $T$, this will imply that $T(a_1v_1+a_2v_2+\ldots+a_nv_n)=0$. Then use the fact that $T$ is one one ($T:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is onto and so its one one also). So you will get $a_1v_1+a_2v_2+\ldots+a_nv_n=0$. Since $\lbrace v_1,v_2,\ldots, v_n\rbrace$ is a basis so $a_1=a_2=\ldots=a_n=0$. Hence $T(v_1),T(v_2),\ldots, T(v_n)$ are linearly independent. 
