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Let $A \in \mathbb{R}^{m \times n}$ a matrix. Show that $\mathscr{N}(A^TA) = \mathscr{N}(A)$, where $\mathscr{N}(A)$ is the null space of the matrix $A$.

Showing that if $x \in \mathscr{N}(A)$ then $x \in \mathscr{N}(A^TA)$ is very straightforward. I'm struggling to show that $\mathscr{N}(A^TA) \subseteq \mathscr{N}(A)$. I could do some algebraic manipulation but it wasn't sufficient:

Let $A=\begin{pmatrix} a_{11} & \dots & a_{1n}\\ \vdots & \ddots & \vdots \\ a_{m1} & \dots & a_{mn} \\ \end{pmatrix}= \begin{pmatrix} b_1 \ ; \dots ; \ b_{n} \end{pmatrix}.$

$A^TA=\begin{pmatrix} b_{1}^T \\ \vdots \\ b_{n}^T \end{pmatrix} \begin{pmatrix} b_1 \ ; \dots ; \ b_{n} \end{pmatrix} = \begin{pmatrix} b_{1}^Tb_{1} & \dots & b_{1}^Tb_{n} \\ \vdots & \ddots & \vdots \\ b_{n}^Tb_{1} & \dots & b_{n}^Tb_{n}\end{pmatrix}$

So we have $A^TAx = \begin{pmatrix} x_{1}b_{1}^Tb_{1} + \dots + x_{n}b_{1}^Tb_{n} \\ \vdots \\ x_{1}b_{n}^Tb_{1} + \dots + x_{n}b_{n}^Tb_{n}\end{pmatrix} = \begin{pmatrix} 0 \\ \vdots \\ 0\end{pmatrix}$

From here I have no idea how to show this makes $Ax=0.$

piero
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1 Answers1

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Without huge computation by assuming general matrix A you can use following steps.

Let,$x \in N(A^tA)$ then, $A^tAx=0$ $x^tA^tAx=0$

$(Ax)^t(Ax)=0$

Let,$Ax=b$ then, $b^tb$=0

$b$ is real vector in $R^n$

So, b=0 & hence, $Ax=0$ which implies $x \in N(A)$

Nope
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