Find a 3 x 3 orthogonal matrix

Question

Find a 3x3 orthogonal matrix whose first column is $[\frac1{\sqrt{3}},{- \frac 1{\sqrt{3}}}, \frac1{\sqrt{3}}]$.

I was thinking this has to do with: ${Q}^T•Q=I_{n}$, but I still don't understand how to go about this.

Answer 1

You need to find an orthonormal basis of $\mathbb{R}^3$ whose first vector is the vector $v_1 = \left( \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)^T$ given to you. This can be done in several ways:

1. Complete $v_1$ arbitrary to a basis $v_1,v_2,v_3$ of $\mathbb{R}^3$ and perform Gram-Schmidt to get $v_1,v_2',v_3'$. Then you can take $Q = (v_1, v_2, v_3)$.
2. Find two linearly independent vectors $v_2,v_3$ that are orthogonal to $v_1$ and perform Gram-Schmidt on them to get $v_2',v_3'$. Then you can take $Q = (v_1,v_2,v_3)$.

Let me demonstrate the second method. Since normalization is not important for orthogonality, we need to find two linearly independent vectors that are orthogonal to $(1,-1,1)^T$. Just by playing with vectors whose entries are $\pm 1$, we can get two such vectors

$$ v_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, v_3 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}. $$

If we perform Gram-Schmidt on $(v_2,v_3)$ we get

$$ v_2' = \frac{v_2}{\| v_2 \|} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \\ v_3' = \frac{v_3 - \left< v_3, v_2 \right> v_2}{\| v_3 - \left< v_3, v_2 \right> v_2 \|} = \frac{\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} - \frac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}}{\dots} = \frac{\begin{pmatrix} -\frac{1}{2} \\ \frac{1}{2} \\ 1 \end{pmatrix}}{\| \begin{pmatrix} -\frac{1}{2} \\ \frac{1}{2} \\ 1 \end{pmatrix} \|} = \begin{pmatrix} -\frac{\sqrt{3}}{\sqrt{10}} \\ \frac{\sqrt{3}}{\sqrt{10}} \\ \frac{\sqrt{4}}{\sqrt{10}} \end{pmatrix} $$

and so we can take $Q$ to be

$$ Q = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & -\frac{\sqrt{3}}{\sqrt{10}} \\ -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{\sqrt{3}}{\sqrt{10}} \\ \frac{1}{\sqrt{3}} & 0 & \frac{\sqrt{4}}{\sqrt{10}} \end{pmatrix}. $$

Answer 2

In this case we can also guess $3$ vectors without calling for Gram-Schmidt by noticing

$\begin{pmatrix} 1 \\ -1 \\ 1\end{pmatrix}\text{ , }\begin{pmatrix} 1 \\ 2 \\ 1\end{pmatrix}\text{ and }\begin{pmatrix} 1 \\ 0 \\ -1\end{pmatrix}\text{ are orthogonal.}$

By keeping first and third coordinate of $v_1,v_2$ equal, we were able to find an easy $v_3$.

Normalizing this gives $Q=\frac 1{\sqrt{6}}\begin{pmatrix} \sqrt{2} & 1 & \sqrt{3} \\ -\sqrt{2} & 2 & 0 \\ \sqrt{2} & 1 & -\sqrt{3} \end{pmatrix}$