Projection and the Orthogonal Projection

Question

Given $$ P = \left \{ (a,b,c,d) \in \mathbb R^4 \mid a + b + c + d = 0 \right \} $$ find $ P^\perp $.

Am I right if I multiply $$ P^T P = 0$$

$P$ orthogonal is only the zero vector?

Answer 1

$P$ is three-dimensional. It's a linear subspace of a four-dimensional space governed by one equation, so its dimension is four - one = three. Its orthogonal complement will be one-dimensional.

A vector $\begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix}$ orthogonal to $P$ satisfies $ax + by + cz + dw = 0$ for all $(a,b,c,d) \in P$. Picking three basis vectors for $P$ will yield you three linear equations for a four-dimensional vector, which you could then solve.

Example: pick $$p_1 = \begin{bmatrix} 1 \\ -1 \\ 0 \\ 0 \end{bmatrix}, p_2 = \begin{bmatrix} 1 \\ 0 \\ -1 \\ 0 \end{bmatrix}, p_3 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ -1 \end{bmatrix}$$

to get the linear system

$$\begin{bmatrix} 1 & -1 & 0 & 0 \\ 1 & 0 & -1 & 0 \\ 1 & 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

and the rest is just linear algebra, which I assume you know.

Answer 2

Since $\dim P=3$, $\dim P^\perp=1$. Actually, $P^\perp=\mathbb R(1,1,1,1)^T$, since, by definition, $P=\bigl((1,1,1,1)^T\bigr)^\perp$.