Find Orthogonal Matrix with 1st Row (1/3,2/3,2/3)

[gysush]

Find an orthogonal matrix whose first row is (1/3,2/3,2/3)

I know orthogonal matrix A satisfies A*A' = I, where A' is the transpose of A and I is identity matrix.

Let A = 1/3*{{1,2,3},{a,b,c},{d,e,f}} where a,b,c,d,e,f elements of R
A'= 1/3*{{1,a,d},{2,b,e},{2,c,f}}

We can obtain 6 equations for 6 unknowns by equation with the identity matrix.
Plugging into mathematica yields

Solve[{a*a + b*b + c*c == 3, a + 2*b + 2*c == 0,
d*d + e*e + f*f == 3, d + 2*c + 2*f == 0, a*d + b*e + c*f == 0,
1 + a*a + d*d == 3} , {a, b, c, d, e, f}]

And the result is highly nonsensical.
What is the problem here? I also know that for orthogonal matrix we can view the columns as an orthonormal set or we could view the rows as an orthonormal set. I.e...a^2 + b^2 + c^2 =3...or 1 + a^2 + d^2 =3...where the RHS of the eq is 3 instead of 1 since the matrix is reduce by a factor of 1/3.

Any help appreciated.
Thanks.

 

[gysush]

I guessed the solution by just trying to make the the squares of the components of the columns equal to 1. Then multiplied the transpose and made terms negative such that they equal 0 when appropriate. So I got that A = 1/3*{{1,2,2},{2,1,-2},{2,-2,1}}
This satisfies A*A' = I. However, what would be a more robust/systematic way? Also, why did my previous method fail?

 

[D H]

Your solution does not satisfy A*A^T = I. You are however close to finding a solution. What's wrong, and how can you fix it?

The problem with your initial approach is that you had false expectations. You implicitly expected to find a set of equations that would yield the solution. The orthogonality conditions give have three equations in six unknowns. That is an underspecified system of equations. There is no one solution. There are instead an infinite number of solutions.

So, how could you have solved this systematically? You have one unit vector. The null space of this vector is a plane. All you have to do is pick any unit vector on this plane as your second unit vector. The third unit vector is (almost) fully constrained. If you want a special orthogonal matrix, the third unit vector is fully constrained.

 

[gysush]

How does my solution not satisfy?

http://www.wolframalpha.com/input/?i=1/3*1/3*{{1,2,2},{2,1,-2},{2,-2,1}}.{{1,2,2},{2,1,-2},{2,-2,1}}

 

[D H]

Sorry, I didn't see the 1/3 in front of the whole thing. I saw your second row as 2,1,-2 rather than 2/3,1/3,-2/3. So, yes, you do have a solution.

Note that this is also a solution:

\bmatrix<br /> \frac 1 3 &amp; \phantom{-}\,\frac 2 3 &amp; \phantom{-}\,\frac 2 3 \\<br /> \frac 2 {\surd 5} &amp; -\,\frac 1 {\surd 5} &amp; \phantom{-}\,0 \\<br /> \frac 2 {\surd 45} &amp; \phantom{-}\,\frac 4 {\surd 45} &amp; -\,\frac 5 {\surd 45}<br /> \endbmatrix

The one you found is nice and rational; the above is not. There are an uncountable number of solutions.

 

[gysush]

Sorry, hope I didn't mean to sound like I was yelling or questioning your help in my response.
Thank you for your response about the plane! I can see what is happening better now.