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.

  • 1
    $\begingroup$ Check that $\ker A = \{0\}$, hence $A^T A$ is invertible. $\endgroup$ – copper.hat Oct 19 '16 at 5:35

For any nonzero vector $v$ in $\Bbb R^m$, you have


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).

  • $\begingroup$ I think you mean $A^tA$ is pos. def., etc. $\endgroup$ – Gerry Myerson Oct 19 '16 at 6:28
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    $\begingroup$ @GerryMyerson Of course you are right, it's the purpose of the proof :) Corrected typo, thanks to have pointed that. $\endgroup$ – Jean-Claude Arbaut Oct 19 '16 at 6:37

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