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Problem: Let $\mathbb{R}^3$ be an Euclidean space. Find an orthonormal basis for the subspace $W$ defined as $x + 2y-z = 0$.

Attempt at solution: So this is a plane in $\mathbb{R}^3$, so I guess I would need two vectors to span this? I did the following. Pick two arbitrary vectors that lie in this plane, for example $w_1 = (1,1,3)$ and $w_2 = (1,0,1)$. Then should I apply the Gram-Schmidt process to these?

That would get me: \begin{align*} v_1 &= w_1 \\ v_2 &= w_2 - \frac{\langle w_2, v_1 \rangle}{\langle v_1, v_1} v_1 \end{align*} Then we would have $v_1 = (1,1,3)$ and $v_2 = (7/11, -4/11, -1/11)$. Now I should normalize these and I'm done?

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    $\begingroup$ Yes, but the results will be simpler if you set $w_1=(1,0,1)$. $\endgroup$ – Bernard Jul 27 '15 at 18:18
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You need two arbitrary linearly independent vectors that lie in the plane. You chose two such vectors, so you've done fine there. You could systematically find such a set in the usual way that you would find the kernel to a linear transformation (i.e. setting "free variables" to certain values etc.).

And yes, everything else that you've done is perfect. Just normalize the vectors and you're all set!

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A simple way of getting independent vectors in this plane is to note that if $ x+ 2y- z= 0$, then $z= x- 2y$ so any vector in this plane is of the form $\begin{pmatrix} x \\ y \\ z\end{pmatrix}= \begin{pmatrix}x \\ y \\ x- 2y\end{pmatrix}$. Taking $x= 1, y= 0$, that is $\begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix}$. Taking $x= 0, y= 1$, this is $\begin{pmatrix} 0 \\ 1 \\ -2\end{pmatrix}$.

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