# Normal equation system

Let $A$ a $n \times m$ matrix with $r(A)=m \le n$ and $b \in R^n$.

How can I prove that the normal equation system $$A^TAx=A^Tb$$ has unique solution denoted by $u$.

If someone can help please.I do not know how to proceed this problem. Thanks for your time and help.

• Check that $\ker A = \{0\}$, hence $A^T A$ is invertible. – copper.hat Oct 19 '16 at 5:35

For any nonzero vector $v$ in $\Bbb R^m$, you have
$$v^TA^TAv=(Av)^T(Av)=||Av||_2^2$$
Since $A$ has rank $m$ by hypothesis (that is, full column rank), $Av$ can't be zero, hence $||Av||_2^2\neq0$ for all $v\neq0$, and your matrix $A^TA$ is positive definite (hence invertible).
• I think you mean $A^tA$ is pos. def., etc. – Gerry Myerson Oct 19 '16 at 6:28