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In a part of proof for least squares problems it is written:

We note that if $A^{*}A$ is singular, then $A^{*}Ax=0$ for some nonzero $x$, implying $x^{*}A^{*}Ax=0$(based on theorem 1).

Theorem 1: Given $A \in C^{m \times n}$ with $m \geq n$ show that $A^{*}A$ is non-singular if and only if $A$ has full rank.

Thus $Ax = 0$, which implies that A is rank-deficient.

I kind of understood this proof for the theorem $1$. But I don't know if I misunderstood something. From $x^{*}A^{*}Ax=0$, can we really conclude $Ax=0$? How?

Thank you.

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$$x^{*}A^{*}Ax=0$$

implies that $$\left\|Ax\right\|^2=0$$

Hence we have $Ax=0$.

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  • $\begingroup$ Thanks, I forgot about this! $\endgroup$ – Crimson Aug 22 '16 at 15:27
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Let $z=Ax$,then

$0=x^{*}A^{*}Ax=z\cdot z=\|z\|^2$

where the "$\cdot $" means the inner product.

And this implies $z=0$.

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    $\begingroup$ Your notation seems a bit mixed up: in a complex vector space, the inner product would already be $z^* z$, so there is no need for the conjugation in $\bar{z}$. $\endgroup$ – Erick Wong Aug 22 '16 at 15:40

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