# Finding eigenvectors of a rank one projection matrix

From the unit vector $$u=\left(\frac{1}{6},\frac{1}{6},\frac{3}6,\frac{5}6\right)$$ construct the rank one projection matrix $$P=uu^t.$$

• a) Show that if $P=uu^t$ , then $u$ is an eigenvector with $\lambda=1$.
• b) If $v$ is perpendicular to $u$, show that $Pv=0$, then $\lambda=0$
• c) Find three independent eigenvectors of $P$ all with eigenvalue of $\lambda=0$.
-
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Julian Kuelshammer Nov 20 '12 at 20:07

1. $Pu$ is an eigenvector means that $Pu = \lambda$. We have $Pu = uu^t u$, which equals what?
2. $v$ perpendicular to $u$ means $u^t v = 0$. Now compute $Pv$.
3. By using the above problem, finding three vectors $v_1, v_2, v_3$ such that $Pv_i = 0$ for each $i$ is the same as finding three vectors each perpendicular to $u$. See if you can set up an equation to solve for all vectors perpendicular to $u$.