Let \(A\) be an \(m \times n\) matrix. Use the steps below to show that a vector \(x\) in \({\mathbb{R}^n}\) satisfies \(Ax = 0\) if and only if \({A^T}Ax = 0\). This will show that \({\rm{Nul}}\,A = {\rm{Nul}}\,{A^T}A\).

1. Show that if \(Ax = 0\), then \({A^T}Ax = 0\).
2. Suppose \({A^T}Ax = 0\). Explain why \({x^T}{A^T}Ax = 0\),and use this to show that \(Ax = 0\).

Given that \(Ax = 0\). Multiply both sides by \({A^T}\) in the given equation and solve as follows:

\(\begin{aligned}{}{A^T}\left( {Ax} \right) &= {A^T}\left( 0 \right)\\{A^T}Ax & = 0\\\left( {{A^T}A} \right)x & = 0\end{aligned}\)

The above calculation shows that \(x \in {\rm{Nul}}\,{A^T}A\). Hence, \({\rm{Nul}}\,A\) is a subsetof \({\rm{Nul}}\,{A^T}A\).

Suppose \({A^T}Ax = 0\). Simplify \({x^T}{A^T}Ax\) as follows:

\(\begin{aligned}{}{x^T}{A^T}Ax & = {x^T}\left( {{A^T}Ax} \right)\\ & = {x^T}\left( {\left( {{A^T}A} \right)x} \right)\\ & = {x^T}\left( 0 \right)\\ & = 0\end{aligned}\)

Since, \({x^T}{A^T}Ax = {\left( {Ax} \right)^T}Ax\). So, \({\left( {Ax} \right)^T}Ax = 0\).

The above calculation implies that \({\left\| {Ax} \right\|^2} = 0\). This is possible if \(Ax = 0\).

This shows that \({\rm{Nul}}\,{A^T}A\)is a subset of \({\rm{Nul}}\,A\).

From part (a), we have \({\rm{Nul}}\,A\) is asubset of \({\rm{Nul}}\,{A^T}A\) and from part (b), we have \({\rm{Nul}}\,{A^T}A\) is a subsetof \({\rm{Nul}}\,A\). Hence, \({\rm{Nul}}\,A = {\rm{Nul}}\,{A^T}A\).