Let $P:R4→R4$ be the orthogonal projection onto the plane
$ W={−(y+2z+t)=0}$
(That is, the projection parallel to the normal vector $(0,−1,−2,−1)$.)
Find the matrix $M^{E}_{E} (P)$ of the projection $P$ relative to the standard basis of $R4$.
$M^{E}_{E} (P)$ = ...
So, I'm kinda new to the topic and confused about this and don't know how to do it. Can someone help me with this.
The simplest way to compute this matrix is to consider how you would normally compute this projection. Let $v$ denote the normal vector $v = (0,-1,-2,-1)$. For a vector $x$, the projection onto the orthogonal complement $W^\perp$ of the hyperplane is given by $$ \operatorname{proj}_{v}(x) = \frac{v^Tx}{v^Tv}\cdot v. $$ So, the projection of $x$ onto $W$ is given by $x - \operatorname{proj}_{v}(x)$. Our goal is to produce a matrix $P$ for which $$ Px = x - \operatorname{proj}_{v}(x) = x - \frac{v^Tx}{v^Tv}\cdot v. $$ A nice way to find this matrix $P$ is to factor the right-hand side into the form $Mx$ for some matrix $M$. In particular, note that $$ x - \frac{v^Tx}{v^Tv}\cdot v = x - \frac 1{v^Tv}v\cdot v^Tx = x - \frac 1{v^Tv}vv^T x = \left(I - \frac{vv^T}{v^Tv} \right)x, $$ where $I$ dentoes the identity matrix. So, our matrix $P$ should be given by $P = I - \frac{vv^T}{v^Tv}.$ If you compute this correctly, you should end up with the answer $$ P = \pmatrix{1 & 0 & 0 & 0\\ 0 & \frac{5}{6} & -\frac{1}{3} & -\frac{1}{6}\\ 0 & -\frac{1}{3} & \frac{1}{3} & -\frac{1}{3}\\ 0 & -\frac{1}{6} & -\frac{1}{3} & \frac{5}{6}}. $$