# Orthonormal basis

• Jun 12th 2009, 05:30 PM
Morgan
Orthonormal basis
Find an orthonormal basis for the vector subspace of $\mathbb R^3,\,\mathbb W=\{(x,y,z)\in\mathbb R^3|x+y+z=0\}.$

Sketch of work:

Spoiler:

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

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

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

well, here's where i dunno what to do, don't know how to get the orthonormal basis for the subspace.
• Jun 12th 2009, 08:13 PM
NonCommAlg
Quote:

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

Sketch of work:

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

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

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

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:

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

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