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So clearly the kernel of $A$ is contained within the kernel of $A^TA$, since

$$\forall x\in \ker A\Rightarrow Ax=0\Rightarrow (A^TA)x=A^T(Ax) = A^T(0) = 0\Rightarrow x\in \ker (A^TA)$$

Now we need to show that the kernel of $A^TA$ is contained within the kernel of $A$. So suppose we have an arbitrary $x \in \ker(A^TA)$, then we have $(A^TA)x = 0$.

How can we show that $Ax=0$?

MathFail
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Parker
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4 Answers4

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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 $\mathbb{R}^n$.

Travis Willse
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    Is it possible to solve this supposing that inner product spaces haven't been covered in the class yet? – Parker Nov 17 '14 at 04:56
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    One can reword the observation by saying that $\vec{b}^T \vec{b} = b_1^2 + \cdots + b_n^2$ is the sum of the squares of the (real) entries of $\vec{b}$. If this sum is zero, then each entry $b_i$ must be so, and so $\vec{b}$ is itself zero. (Of course, this feature is exactly what makes $b \mapsto \sqrt{\vec{b}^T \vec{b}}$ a norm.) – Travis Willse Nov 17 '14 at 05:01
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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.

askuyue
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  • When I ask your question,there is nobody ask,but after i deliver my answer,there is a good answer which is similar with mine. – askuyue Nov 17 '14 at 05:04
2

If $x \in \ker(A^TA)$, $$\Rightarrow A^TAx=0\Rightarrow Ax\in\ker A^T\tag{1}$$

Also, we have $$Ax\in\text{Im} A\tag{2}$$

(1), (2) tell us: $$\Rightarrow Ax\in \ker A^T\cap\text{Im} A$$

We know: $$\ker A^T\perp\text{Im} A \Rightarrow \ker A^T\cap\text{Im} A=\{0\}$$

Therefore,

$$Ax=0$$

MathFail
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0

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 $\vec{0}$.

Parker
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