How do you prove basis associated with col/row space? $A$ is $m\times n$ matrix with $c_1, ...,c_n$. If $\mathrm{rank}\, A = n$, how would you show that {$A^Tc_1,...,A^Tc_n$} is a basis of $ℝ^n$? 
I think that in this case, since there are $n$ vectors and the set is in $ℝ^n$, we only need to prove the linear independence, is it correct? 
Because $\mathrm{rank}\, A = n$ and there are $n$ vectors in $A$, we know that $a_1c_1+...+a_nc_n \implies a_1=...a_n=0$. Using this equation we get $a_1A^Tc_1+...+a_nA^Tc_n=0 \implies A^T(a_1c_1+...+a_nc_n)=0\implies a_1=...=a_n=0$ 
I feel there are many errors I have made or I am just going to the wrong direction. Could anybody help?
 A: You can prove, more generally, that $A$ and $A^T\!A$ have the same rank, because they have the same null space.


*

*If $x\in N(A)$, then $Ax=0$, so $A^T\!Ax=0$; therefore $x\in N(A^T\!A)$.

*If $x\in N(A^T\!A)$, then $A^T\!Ax=0$, so $0=x^T\!A^T\!Ax=(Ax)^T\!(Ax)$, hence $Ax=0$; therefore $x\in N(A)$.


Since both matrices have $n$ columns, the rank nullity theorem says that
$$
\operatorname{rank}A^T\!A=
n-\dim N(A^T\!A)=
n-\dim N(A)=
\operatorname{rank}A
$$
In particular, if $A$ has rank $n$, also $A^T\!A$ has rank $n$. Thus the columns of $A^T\!A$ form a basis for $\mathbb{R}^n$. The set of columns of $A^T\!A$ can be written as $\{A^T\!c_1,A^T\!c_2,\dots,A^T\!c_n\}$ where $c_1,c_2,\dots,c_n$ are the columns of $A$.

What are you missing? The path you chose is good. Let's see how you can complete the proof. Suppose
$$
a_1A^T\!c_1+a_2A^T\!c_2+\dots+a_nA^T\!c_n=0
$$
Then also
$$
A^T\!(a_1c_1+a_2c_2+\dots+a_nc_n)=0
$$
Now write
$$
a_1c_1+a_2c_2+\dots+a_nc_n=
A\begin{bmatrix}a_1\\a_2\\\vdots\\a_n\end{bmatrix}=Av
$$
This means $A^TAv=0$, so also $v^T\!A^T\!Av=0$ and therefore $(Av)^T\!(Av)=0$, so that $Av=0$, which can be rewritten
$$
a_1c_1+a_2c_2+\dots+a_nc_n=0
$$
and the conclusion follows.

Note the similarities between the two proofs. In both I used the fact that if $x$ is a column vector in $\mathbb{R}^n$ and $x^T\!x=0$, then $x=0$.
A: You are on the correct path, but you missed a step. You need to justify why if $A^T(a_1c_1 + \ldots + a_n c_n) = 0$ then $a_1c_1 + \ldots + a_n c_n = 0$ and then use the independence of columns $c_1, \ldots, c_n$ of $A$.
To justify the relevant step, use the fact that the column space of $A$ is orthogonal to the null space of $A^T$.
