What you can conclude about linear transformation

$Q:\mathbb R^n\to \mathbb R^n$ $Q=I-2uu^T$ where is $u\in \mathbb R^n$ 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 $Q^2\not=Q$, 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 $x\in \mathbb R^n$,but Is there more important to say?

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

• Do you have something to send me about reflection, something with theorems and proofs? – Marko Škorić Sep 6 '18 at 10:12

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

• I know where I made mistake I saw $u^tu$ not $uu^t$,I hate when that is happens – Marko Škorić Sep 6 '18 at 10:26