c0mpatible system $A^TAx=A^Tb$

Question

Let $A\in\mathbb{R}^{n\times n}$ be a singular matrix. Prove that the system $$A^TAx=A^Tb$$ is compatible for any $b\in\mathbb{R}^n$.

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I want to prove that $A^Tb\in Ran(A^TA)$,i.e. $A^Tb\bot Ker(A^TA)$

Answer 1

Equivalent statements:

• $x\in\mathrm{ker}(A^TA)$,
• $Ax=0$.

Proof: $x\in\mathrm{ker}(A^TA)$ implies that $x^TA^TAx=(Ax)^T(Ax)=\|Ax\|_2^2=0$. Hence $Ax=0$. The other direction is trivial.

Hence $A^Tb\perp\mathrm{ker}(A^TA)$, because $y^TA^Tb=(Ay)^Tb=0^Tb=0$ for any $y\in\mathrm{ker}(A^TA)$.