# Orthogonal projection and orthogonal complements onto a plane

Let $V$ be a subspace of $\mathbb{R}^n$ with $V \ne \mathbb{R}^n$ and $V\ne \{0\}$. Let $A$ be the matrix of the linear transformation $\text{proj}_V : \mathbb{R}^n\to\mathbb{R}^n$ that is the projection onto $V$.

Calculate $A$ in the case where $n=3$ and $V$ is given by the equation $x+y+z=0$.

I am not really sure how to approach this question. I am aware that the normal vector will be $(1,1,1)$. And I know that the orthogonal projection onto the plane $V$ given by the equation $x+y+z=0$ is equal to the identity minus the orthogonal projection onto the orthogonal complement. I believe the orthogonal complement will be in the span of the normal vector (not entirley sure why?) but I am also unsure how to find the orthogonal complement.

You are right: $(1,1,1)$ is orthogonal to $V$. Therefore, $A.(1,1,1)=(0,0,0)$. Now, consider the vectors $(1,-1,0)$ and $(1,0,-1)$. Since they both belong to $V$, you must have $A.(1,-1,0)=(1,-1,0)$ and $A.(1,0,-1)=(1,0,-1)$.

Now, since$$(1,0,0)=\frac13(1,1,1)+\frac13(1,-1,0)+\frac13(1,0,-1),$$you must have$$A.(1,0,0)=\frac13(1,-1,0)+\frac13(1,0,-1)=\left(\frac23,-\frac13,-\frac13\right).$$So, the entries of the first column of the matrix of $\operatorname{proj}_V$ with respect to the standard basis will be $\frac23$, $-\frac13$ and $-\frac13$. Can you take it from here?

• Where are you getting the 1/3 from? – Maths student Jul 27 '18 at 20:50
• From solving the system $(1,0,0)=\alpha(1,1,1)+\beta(1,-1,0)+\gamma(1,0,-1)$. – José Carlos Santos Jul 27 '18 at 21:11

Hint:

The projection of a vector $\vec u$ onto the normal to the plane $x+y+z=0$, with normal vector $\vec n=(1,1,1)$, is given by the formula: $$p_{\vec n}(\vec u)=\vec u-\frac{\vec n\cdot\vec u}{\vec n\cdot\vec n}\,\vec n.$$ Apply this formula to each vector of the canonical basis to find the column vectors of the projection matrix.

As you say, $u:=(1,1,1)^T$ is a normal vector to the plane, thus it will span the orthogonal complement.

Its projection is given by $\varphi:v\mapsto\frac{u^Tv}{u^Tu}u$.
You can check that it fixes $u$, but gives $0$ whenever $v\perp u$.
To obtain its matrix in the standard basis, just calculate $\varphi(e_i)$ for the standard basis $e_1,e_2,e_3$.

$\left\{\frac1{\sqrt{2}}(1,-1,0), \frac1{\sqrt{6}}(1,1,-2)\right\}$ is an orthonormal basis for $V$ so the projection is given by

\begin{align} P(x,y,z) &= \left\langle (x,y,z), \frac1{\sqrt{2}}(1,-1,0)\right\rangle\frac1{\sqrt{2}}(1,-1,0)+\left\langle (x,y,z), \frac1{\sqrt{6}}(1,1,-2)\right\rangle\frac1{\sqrt{6}}(1,1,-2) \\ &= \frac12 (x-y, -x+y,0) + \frac16(x+y-2z,x+y-2z,-x-y+4z)\\ &= \frac16 (4x-2y-2z, -2x+4y-2z, -2x-2y+4z) \end{align}

so $$A = \frac16\begin{bmatrix} 4 & -2 & -2 \\ -2 & 4 & -2 \\ -2 & -2 & 4 \end{bmatrix}$$

The orthogonal projection of $(x,y,z)$ onto the plane through the origin with normal $(1,1,1)$ is the unique $(x',y',z')$ on the plane such that $$((x,y,z)-(x',y',z')) = c(1,1,1), \\ x'+y'+z'=0.$$ Dotting the first equation with $(1,1,1)$ and using $x'+y'+z'=0$ gives $$c = \frac{1}{3}(x+y+z).$$

So the projection of $(x,y,z)$ onto the given plane is

\begin{align} (x',y',z') &= (x,y,z)-\frac{1}{3}(x+y+z)(1,1,1) \\ &= \frac{1}{3}(2x-y-z,-x+2y-z,-x-y+2z). \end{align}