# Given two orthogonal (orthonomal also) vectors $(1,0,0)^T,(0,\frac{1}{\sqrt{5}},\frac{-2}{\sqrt{5}})^T$

Given two orthogonal (orthonomal also) vectors $$(1,0,0)^T,(0,\frac{1}{\sqrt{5}},\frac{-2}{\sqrt{5}})^T$$

I want to find a third vector $$q_3$$ such that they become all orthogonal to each other. Of course one should directly apply the gram-schimdt process. But by my books definition is as follows :

Suppose $$(\mathbf{u}_1, \mathbf{u}_2, \dots \mathbf{u}_n)$$ is a basis for a finite dimensional inner product space $$V$$. Let $$\begin{eqnarray*} \mathbf{v}_1 &=& \mathbf{u}_1,\\ \mathbf{v}_2 &=& \mathbf{u}_2 -\frac{\langle \mathbf{u}_2,\mathbf{v}_1\rangle}{\langle \mathbf{v}_1,\mathbf{v}_1\rangle}\mathbf{v}_1,\\ \mathbf{v}_3 &=& \mathbf{u}_3 -\frac{\langle \mathbf{u}_3,\mathbf{v}_1\rangle}{\langle \mathbf{v}_1,\mathbf{v}_1\rangle}\mathbf{v}_1-\frac{\langle \mathbf{u}_3,\mathbf{v}_2\rangle}{\langle \mathbf{v}_2,\mathbf{v}_2\rangle}\mathbf{v}_2,\\ \vdots\\ \mathbf{v}_n &=& \mathbf{u}_n -\frac{\langle \mathbf{u}_n,\mathbf{v}_1\rangle}{\langle \mathbf{v}_1,\mathbf{v}_1\rangle}\mathbf{v}_1-\frac{\langle \mathbf{u}_n,\mathbf{v}_2\rangle}{\langle \mathbf{v}_2,\mathbf{v}_2\rangle}\mathbf{v}_2-\cdots-\frac{\langle \mathbf{u}_n,\mathbf{v}_{n-1}\rangle}{\langle \mathbf{v}_{n-1},\mathbf{v}_{n-1}\rangle}\mathbf{v}_{n-1} \end{eqnarray*}$$ Then $$\{\mathbf{v}_1, \mathbf{v}_2,\dots,\mathbf{v}_n\}$$ is an orthogonal basis for $$V$$ .

So a question popped out, my aim is to find $$\mathbf{v}_3$$, since I already know what my $$\mathbf{v}_1,\mathbf{v}_2$$ are, It suffices to just choose my $$\mathbf{u}_3$$, so when choosing my $$\mathbf{u}_3$$, should I just choose $$\mathbf{u}_3$$ to be any vector from a basis that forms $$\mathbb{R}^3$$ (avoiding $$(1,0,0)^T$$ in this case) or is there a systematic way to choose $$\mathbf{u}_3$$?

• The idea of the Gram-Schmidt process is that any basis can be turned into an orthogonal basis through this process and so it would be natural to choose a linearly independent vector $\textbf{u}_3$ at "random". However in Your case You might just set $\textbf{u}_3=\textbf{u}_1\times \textbf{u}_2$ that is already orthogonal – Peter Melech Sep 23 '18 at 10:03

I would introduce a new vector $$(a,b,c), a,b,c \in \mathbb{R}$$, and start finding the conditions for which is linearly independent with the others, i.e. $$\det \left( \begin{matrix} 1& 0 & 0\\ 0& 1/\sqrt{5}& -2/\sqrt{5} \\ a & b& c \end{matrix} \right) \neq 0.$$ As you can compute, this gives $$c + 2b \neq 0$$. Under this condition, you can apply the Gram-Schmidt process and find an orthonormal basis.