Proving invertibility of $A^{T}A$

Question

Take the matrix $A$, which is $m\times n$ and of rank $n$. Hence, a full column-rank matrix.

I need to show that $N(A^{T}A) = N(A)$, and deduce that $A^{T}A$ is invertible.

Since A is a full column-rank matrix, I see that $dim \space N(A) = n-r = n-n =0$, since $r=n$.

Howver, I'm not sure how to characterize the null-space of $N(A^{T}A)$ in order to arrive at $N(A^{T}A) = N(A)$.

Once this is done, I see how $N(A) = \vec{0}$ implies the invertibility of $A^{T}A$.

Answer 1

We need to show that $N(A) = N(A^{T}A)$.

Take $Av=0$. Then $A^{T}(Av) = (A^{T}A)v=0$, so we have that $v \in N(A) \rightarrow v \in N(A^{T}A)$ and $N(A) \subseteq N(A^{T}A)$.

In the other direction, starting with $v \in N(A^{T}A)$:

$A^{T}Av=0 \rightarrow (v^{T}A^{T})(Av) =0$

Since $(v^{T}A^{T})(Av) = ||Av||^{2}$, and the only vector $v$ in $\mathbb{R}$ which has norm $0$ is $\vec{0}$, we have that $v \in N(A^{T}A) \rightarrow v \in N(A)$ and $N(A^{T}A) \subseteq N(A)$.

So $N(A^{T}A) = N(A) = 0$ and $A^{T}A$ is invertible—since A is full rank.

Answer 2

This is not true for matrices over a general coefficient field.

For example, take coefficient field $\Bbb C$ and $A = \begin{pmatrix}1 \\ i\end{pmatrix}$. Then $A$ has full column-rank but $A^TA = \begin{pmatrix}0\end{pmatrix}$ is not invertible.

This is true for matrices over $\Bbb R$, as @lmaosome pointed out in the comment.

Namely, $A^TAv = 0$ implies $0 = v^TA^TAv = |Av|^2$. Over $\Bbb R$, the only vector that has norm $0$ is the zero vector. Thus we have $Av = 0$.