Prove that $\det(A^{T}A) \neq 0$ How to prove that $\det(A^{T}A) \neq 0$ if coloumns of $A$ are linearly independent, without using Cauchy-Binet formula? $A$ is real matrix.
 A: We are assuming that $A$ is real and $n\times m$. 
Suppose that $\det(A^TA)=0$. Then there exists nonzero $v\in \mathbb R^n$ such that $A^TAv=0$. Then 
$$
0=A^TAv=v^TA^TAv=(Av)^TAv.
$$
For any real vector $w$, if $w^Tw=0$, then $w=0$. Thus, $Av=0$. But this is exactly, if $C_1,\ldots,C_m$ are the columns of $A$,
$$
0=v_1C_1+v_2C_2+\cdots+v_mC_m.
$$
So the columns of $A$ are linearly dependent. 
If follows that if the columns of $A$ are linearly independent, then $\det(A^TA)\ne0$. 
A: We can prove a much more general (and useful) result.

Theorem. The (real) matrices $A$ and $A^TA$ have the same null space, hence the same rank.

(I'll use $N(B)$ to denote the null space of $B$: $N(B)=\{v:Bv=0\}$.)
Proof. It is obvious that $N(A)\subseteq N(A^TA)$. Suppose $v\in N(A^TA)$; then $A^TAv=0$ implies $v^TA^TAv=0$, so $(Av)^T(Av)=0$, hence $Av=0$.
If $A$ is $m\times n$, then the rank of $A$ is $n-\dim N(A)$ by the rank-nullity theorem. Similarly the rank of $A^TA$ is $n-\dim N(A^TA)$, thus proving the claim.$~\square$
In the case when the columns of $A$ are linearly independent, then the rank of $A$ equals the number of columns; therefore the rank of $A^TA$ equals the number of columns. Since $A^TA$ is square, it follows it is invertible, so $\det(A^TA)\ne0$.

Note that this exploits the fact that the matrices are real and the result need not hold over other fields. For a counterexample, consider the complex matrix
$$
A=\begin{bmatrix}1\\i\end{bmatrix}
$$
Then
$$
A^TA=\begin{bmatrix}1 & i\end{bmatrix}\begin{bmatrix}1\\i\end{bmatrix}=[0]
$$
The result holds in ordered fields, where a sum of nonzero squares is nonzero. More generally, over fields where $-1$ is not a sum of squares (because they can be ordered). A counterexample can always be found when $-1$ is a sum of squares: if $-1=a_1^2+\dots+a_n^2$, then the matrix
$$
A=\begin{bmatrix}a_1\\\vdots\\a_n\\1\end{bmatrix}
$$
has $A^TA=0$.
A: Since the columns of $A$ are linearly dependent then $A$ is injective. In fact if $Ae_1,\ldots,Ae_n$ are the columns of $A$ and if $x=\sum_{i=1}^nx_ie_i\in \ker A$ then $0=Ax=\sum_{i=1}^n x_i Ae_i\implies x_i=0\implies x=0$. Now if $\det(A^TA)=0$ then $0$ is an eigenvalue of $A^TA$ and let $x$ an associated eigenvector. Then
$$0=\langle A^TAx,x\rangle=\Vert Ax\Vert^2\implies Ax=0\iff x\in\ker A$$
which is an obvious contradiction.
