Why is $x^T(uu^T)x=(x^Tu)(u^Tx) = (\langle x, u \rangle)(\langle x, u \rangle)=((\langle x, u \rangle))^2 \geq 0$

Question

Given that u is a d x 1 vector such that ||u|| = 1 and for all d x 1 vector x, how can we reason through the equation ? $x^T(uu^T)x=(x^Tu)(u^Tx) = (\langle x, u \rangle)(\langle x, u \rangle)=((\langle x, u \rangle))^2 \geq 0$

In this case, x is a scalar(?). and $(uu^T)$ gives us 1(?) which mean $\sum_1^d$ $x_1\cdot x_1$, Then since ||u|| is 1 given. what can we say about each $x_1$ if x is a scalar, can we even transpose it ?

Answer 1

$x$ and $u$ are vectors. Notice that if $x=(x_1,...,x_n)$, then $$x^Tx=\sum_{i=1}^n x_i^2=\|x\|^2.$$ Then, $$(x^Tu)(u^Tx)=(u^Tx)^T(u^Tx)=\|u^Tx\|^2\geq 0.$$