Show that if a vector $v$ is orthogonal to $u$ then $Pv=0$ being $P=uu^T$

Question

Show that if a vector $v$ is orthogonal to $u$ then $Pv=0$ being $P=uu^T$ and $\displaystyle u=(\frac{1}{6},\frac{1}{6},\frac{3}{6},\frac{5}{6})$.

First I showed that $uu^T=1$, and the hypothesis of this problem tells me that $<v,u>=0$ (since they're orthogonal).

• Note: <u,v> is an inner product.

If $P=1$, why would $Pv=0$ and not lead to $v$?

Answer 1

Note that vectors are denoted by column matrices, not row matrices. So $u=\begin{bmatrix}1/6\\1/6\\3/6\\5/6\end{bmatrix}=\begin{bmatrix}1/6&1/6&3/6&5/6\end{bmatrix}^T$, so your question is missing the transpose.

Now $uu^T$ is not $1$ but a $4\times4$ matrix. Given $\langle u,v\rangle=u^Tv=0$, we have $Pv=(uu^T)v=u(u^Tv)=\vec0$ by associativity of matrix product.