(Algebra Review)** Extend the argument supporting Lemma 2.1 to show that $$rank(X) = rank(X'X)$$.

Use the following steps.

(a) Show that for all $$x \in R^k$$, $$X'Xx = 0$$ implies that $$Xx = 0$$. (HINT: $$Xx = 0$$ if and only if $$\|Xx\| = 0$.)

(b) Because $$Xx = 0$$ also implies $$X'Xx = 0$$, argue that $$Col^\perp(X') = Col^\perp(X'X)$$.

(c) Use Theorem C.11, which states that

$$N = dim[Col(X)] + dim[Col^\perp(X)]$$

and Theorem C.12, which states that

$$rank(X) = dim[Col(X)] = dim[Col(X')] = rank(X')$$

to show that $$rank(X) = rank(X'X)$$. (HINT: Apply Theorem C.11 to X' and X'X.)