1. ## Orthonormal basis

Find an orthonormal basis for the vector subspace of $\displaystyle \mathbb R^3,\,\mathbb W=\{(x,y,z)\in\mathbb R^3|x+y+z=0\}.$

Sketch of work:

Spoiler:

i'll take $\displaystyle x=-y-z,$ so we have that

$\displaystyle \mathbb W=\{(-y-z,y,z)\in\mathbb R^3,\,\forall\,y,z\in\mathbb R\},$ thus $\displaystyle \mathbb W=\langle(-1,1,0),(-1,0,1)\rangle.$

these are linearly independent, so i have a basis for $\displaystyle \mathbb W$ which is $\displaystyle B_{\mathbb W}=\{(-1,1,0),(-1,0,1)\}.$

well, here's where i dunno what to do, don't know how to get the orthonormal basis for the subspace.

2. Originally Posted by Morgan
Find an orthonormal basis for the vector subspace of $\displaystyle \mathbb R^3,\,\mathbb W=\{(x,y,z)\in\mathbb R^3|x+y+z=0\}.$

Sketch of work:

i'll take $\displaystyle x=-y-z,$ so we have that

$\displaystyle \mathbb W=\{(-y-z,y,z)\in\mathbb R^3,\,\forall\,y,z\in\mathbb R\},$ thus $\displaystyle \mathbb W=\langle(-1,1,0),(-1,0,1)\rangle.$

these are linearly independent, so i have a basis for $\displaystyle \mathbb W$ which is $\displaystyle B_{\mathbb W}=\{(-1,1,0),(-1,0,1)\}.$

well, here's where i dunno what to do, don't know how to get the orthonormal basis for the subspace.
first, you don't need to hide your own work! second, you're right so far. now you just need to apply Gram-Schmidt process. let's give a name to the elements of the basis you found:

$\displaystyle u_1=(-1,1,0), \ u_2=(-1,0,1).$ now find $\displaystyle v_1=u_2-\frac{<u_1,u_2>}{<u_1,u_1>}u_1.$ the set $\displaystyle \{u_1,v_1 \}$ is an orthogonal basis for $\displaystyle W.$ so you just need to normalize them, i.e. find $\displaystyle e_1=\frac{u_1}{||u_1||}, \ e_2=\frac{v_1}{||v_1||}.$

then $\displaystyle \{e_1,e_2 \}$ would be an orthonormal basis for $\displaystyle W.$