Orthogonal Matrix with a specific row

Question

I have an assignment with the following question:

Does an Orthogonal Matrix exist such that its first row consists of the

 following values:

($1$/$\sqrt{3}$, $1$/$\sqrt{3}$, $1$/$\sqrt{3}$)

If there is, find one.

I know I can solve this question with the Gram Schmidt algorithm, but it includes a lot of complicated calculations.

Is there any other way to prove this statement without the Gram Schmidt algorithm?

Thanks,

Alan

Answer 1

You need to choose two vectors which are orthogonal to $\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$ and make sure they are also orthogonal to each other.

Vectors orthogonal to $\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$ lie in the plane $x+y+z=0$.

For example choose $(1,-1,0)$ to be the second row of your matrix.

Now you need a third vector which satisfies the equations:

$$x+y+z=0 ~~~ \text{[is orthogonal to}~\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)]$$ $$x-y=0 ~~~ \text{[is orthogonal to}~(1,-1,0)]$$

Any such vector can form the third row of your orthogonal matrix.

Answer 2

i answered a similar question before, but i don't know how to make a link to that.

so i will repeat the way to find a symmetric reflection matrix whose row is your unit vector. reflect $((1,1,1)^T/\sqrt 3$ to $(1,0,0)^T$ on the mirror in the direction of their bisector $a = (1 + \sqrt 3, 1, 1)^T.$ form the matrix $q = 2aa^T/a^Ta - I;$ that will satisfy your requirement.