0
$\begingroup$

$Q:\mathbb R^n\to \mathbb R^n$ $Q=I-2uu^T$ 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?

$\endgroup$
0
$\begingroup$

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 $u$.

$\endgroup$
  • $\begingroup$ Do you have something to send me about reflection, something with theorems and proofs? $\endgroup$ – Marko Škorić Sep 6 '18 at 10:12
1
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.