Determine the orthogonal complement of the subspace of $\mathbb{R}^3$ given by

Question

Determine the orthogonal complement of the subspace of $\mathbb{R}^3$ given by: $$V=\{(x,y,z)^T:x-y=0\}$$ and find an orthonormal basis for it.

My try:

$x-y=0\implies x=y$ $$\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} y \\ y \\ z \end{bmatrix}=y\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}+z\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$ Let $u_1=\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix},u_2=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ $$u_1\cdot u_2=\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}\cdot \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}=0$$So, they are orthogonal.

And to determine the orthonormal basis I need to used Gram-Schmidt right?

Can anyone tell whether I am doing right or wrong?

Answer 1

$V$ is a two-dimensional subspace, spanned by $$ \left[\begin{array}{c}1 \\ 1 \\ 0 \end{array}\right],\;\left[\begin{array}{c}0 \\ 0 \\ 1\end{array}\right]. $$ The orthogonal complement of $V$ is a one-dimensional subspaced, spanned by

$$ \left[\begin{array}{c} 1 \\ -1 \\ 0 \end{array}\right]. $$ So an orthonormal basis of $V^{\perp}$ is $$ \left\{ \left[\begin{array}{c} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{array}\right]\right\} $$

Answer 2

Vectors with $x=y$ belong to $V$. You’ve found a basis for $V$, not for its orthogonal complement.

If you interpret the left-hand side of $V$’s defining equation as a dot product, what is says is that $V$ consists of all vectors that are orthogonal to a particular fixed vector. It should be pretty clear, then, that all vectors orthogonal to $V$ are scalar multiples of this fixed vector. To get an orthonormal basis for $V^\perp$ all you need to do is normalize this vector, which I’ll leave to you to figure out.