$uu^T$ is the standard matrix for the orthogonal projection of $\mathbb{R}^n$ on the subspace spanned by $u$

Let $A$ be an $n\times n$ symmetric matrix, and $u$ an eigenvector of $A$. Why is it true that $\forall x\in\mathbb{R}^n$, $x^{T}uu^{T}(I-uu^T)x=0$? If I'm able to show this is true then I can show that $uu^T$ is the standard matrix for the orthogonal projection of $\mathbb{R}^n$ on the subspace spanned by $u$. Thanks.

1 Answer

$uu^T(I-uu^T)=uu^T-uu^Tuu^T= uu^T- u(u^Tu)u^T= uu^T(1-u\cdot u)$