# Linear algebra:P2 - orthonormal basis of V and $\mathbb{R^3}$

In $\mathbb{R}^3$ with $\underline{N}= \begin{pmatrix} 1\\ 2\\ 1 \end{pmatrix}$
and the subspace $V={\underline{N}^\bot}$.

From Linear Algebra:P1 - Dim(V), linear independence follows:
(a1)The following vectors are linearly independent
(a2)dim(V)=2.
(a3)The following vectors are in V.

$\underline{x}_1= \begin{pmatrix} 0\\ 1/2\\ -1 \end{pmatrix}$ , $\underline{x}_2= \begin{pmatrix} 1\\ 0\\ -1 \end{pmatrix}$ , $\underline{x}_3= \begin{pmatrix} 1\\ -1\\ 1 \end{pmatrix}$

(a4)Construct/built an orthonormal basis {$\underline{w}_1;\underline{w}_2$} of V with $\underline{v}_1$ and $\underline{v}_2$ by using the Gram–Schmidt process.

(a4)
1) $$\underline{v}_1:=\underline{x}_1 \ne \underline{0}$$

$$\underline{v}_1= \begin{pmatrix} 1\\ 2\\ 0 \end{pmatrix}$$

$$\underline{w}_1= \frac{ \underline{v}_1 }{\|\underline{v}_1\|}= \frac{2}{\sqrt{5}} \cdot \begin{pmatrix} 0\\ 1/2\\ -1 \end{pmatrix}$$

2)$$\underline{v}_2:=\underline{x}_2- \frac{<\underline{x}_2;\underline{v}_1>}{<\underline{v}_1;\underline{v}_1>}\cdot \underline{v}_1$$

$$\underline{v}_2:= \begin{pmatrix} 1\\ 0\\ -1 \end{pmatrix}- \frac{\begin{pmatrix} 1\\ 0\\ -1 \end{pmatrix} \cdot \begin{pmatrix} 0\\ 1/2\\ -1 \end{pmatrix}} {\begin{pmatrix} 0\\ 1/2\\ -1 \end{pmatrix} \cdot \begin{pmatrix} 0\\ 1/2\\ -1 \end{pmatrix}} \cdot \begin{pmatrix} 0\\ 1/2\\ -1 \end{pmatrix}$$

$$\underline{v}_2= \begin{pmatrix} 1\\ 0\\ -1 \end{pmatrix} - \frac{4}{5} \cdot \begin{pmatrix} 0\\ 1/2\\ -1 \end{pmatrix} = \begin{pmatrix} 1\\ -2/5\\ -1/5 \end{pmatrix}$$

$$\underline{w}_2= \frac{ \underline{v}_2 }{\|\underline{v}_2\|}= \frac{1}{\sqrt{6}} \cdot \begin{pmatrix} 1\\ -2/5\\ -1/5 \end{pmatrix}$$
So that {$\underline{w}_1;\underline{w}_2$} are an orthonormal basis of V.

(a5)Extend {$\underline{w}_1;\underline{w}_2$} to an orthonormal basis for $\mathbb{R^3}$

My questions are:
Is my approach correct for (a4)?
Have I done any mistake?
For (a5), how can I extend it, shall I take $\underline{x}_3$ and calculate with this one $\underline{w_3}$ by $$\underline{w}_3= \frac{\underline{v}_3}{\|\underline{v}_3\|}$$
or is there another approach, perhaps by using $\underline{N}$, and if it is,
why is it this way?

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To check if you have an orthonormal set, you do the inner product between the elements in your set. If the inner product is 0 and every element has norm 1, then you have an orthonormal set. For your second question, have you heard of the Replacement lemma? A corollary to it is that you can extend a set to be a basis of a vector space V, provided that set is linearly independent. –  noobProgrammer Apr 3 '13 at 0:39