Finding eigenvectors of a rank one projection matrix

Question

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 $\lambda=0$.

Answer 1

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 $u$.