Vtyjkyuk

$Q:mathbb R^nto mathbb R^n$ $Q=I-2uu^T$ where is $uin mathbb R^n$ such that $||u||_2=1$, what you can conclude about linear transformation.

First $Q=I-2uu^T=I-2||u||^2=I-2I=-I$, first Q is not projection matrix because $Q^2not=Q$, but $Q^3=Q$,all eigenvalue is $-1$, it is orthogonal matrix since $Q^TQ=QQ^T=I$, and I know that $mathbb R^n=ker(Q)⊕Im(Q)$, $ker(Q)=0$, Is there something more to say, I mean since this matrix is orthogonal we can say that $||Qx||_2=||x||_2$ for every $xin mathbb R^n$,but Is there more important to say?

Denote by $langle v,wrangle = v^T w$ the usual dot product on $mathbb R^n$. The map you are given acts as
$$
Qv = (I-2uu^T)v = v - 2 langle u, vrangle u.
$$
Since $|u|=1$, this is a reflection at the plane orthogonal to $u$.

Let's be more detailed about $Q^2$ (you got it wrong):
$$
Q^2=(I-2uu^T)(I-2uu^T)=I-2uu^T-2uu^T+4uu^Tuu^T=I
$$
because $uu^Tuu^T=(u^Tu)uu^T=|u|^2uu^T=uu^T$. Since $Q^T=(I-2uu^T)^T=I-2uu^T=Q$, we conclude that $Q=Q^T$ is orthogonal.

Besides, you write $Q=-I$, which is definitely wrong.

An orthogonal matrix is a projection if and only if it is the identity (prove it, no need to go with eigenvalues).

$Q=Q^2$ implies $I=Q^TQ=Q^TQ^2=Q$