- #1
gysush
- 26
- 0
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.
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.