# Rank of $A^T A$ where $A$ has a full column rank

I'm having some trouble proving this: Suppose $A$ is a rectangular $m\times n$ matrix ($m > n$),

If $A$ has a full column rank ($\text{r}(A)=n$) then so does $A^TA$ (= it is invertible)

I read a proof that tries to show that $\mathcal{N}(A)=\{\mathbf 0\}$ implies $\mathcal{N}( A^TA)=\{\mathbf 0\}$, but I can't understand why that is right.

• We use mathjax formatting here. Formatting tips here. – Em. Feb 22 '16 at 19:50
• What is the function N? – Bernard Feb 22 '16 at 19:54
• Take a look at the top answer to this question: math.stackexchange.com/questions/215145/… – timothyg Feb 22 '16 at 20:05
• @user297699 Thanks that helped – idanp Feb 22 '16 at 20:44

Let $x \in \mathcal{N}(A^TA)$. Hence $A^TAx=0_{n \times 1}$, thus $$\|Ax\|^2=x^TA^TAx=x^T0_{n \times 1} = 0,$$ therefore $Ax=0$, thus $x \in \mathcal{N}(A)$, but since $\text{r}(A)=n$ we must have that $\mathcal{N}(A)=\{0\}$, and then $x=0$. So indeed $\mathcal{N}(A^TA)=\{0\}$, which easily implies that $\text{r}(A^TA)=n$ as wanted.

• I was missing the fact that if a space is contained in {0} then it has to be {0} as well. Thanks. – idanp Feb 22 '16 at 20:46