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So clearly the kernel of $A$ is contained within the kernel of $A^TA$, since $$A^T(A\vec{x}) = \vec{0} \Rightarrow A^T(\vec{0}) = \vec{0}$$. Now we need to show that the kernel of $A^TA$ is contained within the kernel of $A$. So suppose we have a $\vec{x} \in \ker(A^TA)$ so that $(A^TA)\vec{x} = \vec{0}$.

How can we show that $A\vec{x}$ must also be equal to $\vec{0}$?

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Hint: Multiplying on the left by $\vec{x}^T$ gives

$${\vec{x}}^T A^T A \vec{x} = 0.$$

This gives $0 = (A \vec{x})^T (A \vec{x}) = \left\vert\left\vert A \vec{x}\right\vert\right\vert^2$, where $||\cdot||$ is the standard norm on $\mathbb{R}^n$.

For any $x \in Ker(A^{T}A)$ ,which means $A^{T}Ax = 0 $,and we have

\begin{equation}

x^{T} A^{T} A x = (Ax)^{T}Ax = 0

\end{equation}.

Hence we must have $Ax = 0$.

Maybe helps.

Given $A^TA\vec{x} = 0$, we have that $A\vec{x}$ is in the image of $A$ as well as in the kernel of $A^T$. Since the kernel of $A^T$ is the orthogonal complement of the image of $A$, we have that $A\vec{x}$ is $\vec{0}$.

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