Prove that for a real matrix $A$, $\ker(A) = \ker(A^TA)$
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 $\vec{0}$?
$\endgroup$3 Answers
$\begingroup$Hint: Multiplying on the left by $\vec{x}^T$ gives $${\vec{x}}^T A^T A \vec{x} = 0.$$
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ 1 $\begingroup$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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