Answer to Question #150985 in Linear Algebra for Ashweta Padhan

Question

Find an orthogonal matrix P whose first row is u1=(1/3, 2/3, 2/3)

Answer 1

Since we need matrix A є M3x3 (F) which is orthogonal, we want orthonormal basis for F³ whose first vector is given row $(1/3; 2/3;2/3)$ (because row of an orthogonal matrix M form an orthonormal basis for Fⁿ).

Let $\gamma = {\{(1/3, 2/3, 2/3), (1,0,0), (0,1,0)}\}$ and it be ordered basis for F³ with given row as first element.

Set $\omega1 = (1/3,2/3,2/3), \omega2 = (1,0,0), \omega3 =(0,1,0)$ and use Gram-Schmidt orthogonalization process to get orthogonal basis for F³ .

$$\upsilon1 = \omega1 = (1/3, 2/3, 2/3)$$

$\upsilon2 = \omega2 - \upsilon1 * <\omega2, \upsilon1>/<\upsilon1, \upsilon1> =$

$= (1,0,0) - (1/3,2/3,2/3)*(1/3)/(1/9+4/9+4/9)$

$=(1-1/9,0-2/9, 0-2/9) = (8/9, -2/9, -2/9)$ $= (1,0,0 ) - 1/3*(1/3,2/3,2/3)$

$\upsilon3 = \omega3 - \upsilon1 * <\omega3, \upsilon1>/<\upsilon1, \upsilon1> -\upsilon2 * <\omega3, \upsilon2>/<\upsilon2, \upsilon2> =$

$= (0,1,0) - (1/3, 2/3, 2/3)*(2/3)/(1/9+4/9+2/9) - (8/9,-2/9,-2/9)*(-2/9)/(64/81+4/81+4/81)$$=(0,1,0) - 2/3 *(1/3,2/3,2/3) +1/4*(8/9, -2/9,-2/9)$ $=$

$= (0 - 2/9+8/36, 1-4/9-2/36, 0 - 4/9-2/36)$

$= (0, 1/2, -1/2)$

$\upsilon1,\upsilon2, \upsilon3$ is now orthogonal basis which only needs to be normalized.

Define $ui = \upsilon i/||\upsilon i|| for$ i=1,2,3

$\upsilon1$ is normal yet, so u1 = $\upsilon1$ = (1/3, 2/3, 2/3)

$u2 = \upsilon2/||\upsilon2||$

$=(8/9,-2/9,-2/9)/\sqrt{\smash[b]{8/9*8/9+(-2)/9*(-2)/9 +(-2)/9*(-2)/9}}$

$= (8/9,-2/9,-2/9)/(2*\sqrt{\smash[b]2}/3)$

$= (2*\sqrt{\smash[b]2}/3, -\sqrt{\smash[b]2}/6, -\sqrt{\smash[b]2}/6)$

$u3 = \upsilon3/||\upsilon3||$

$=(0,1/2,-1/2)/\sqrt{\smash[b]{0+1/2*1/2 +(-1)/2*(-1)/2}}$

$= (0,1/2,-1/2)/(1/\sqrt{\smash[b]2})$

$= (0, \sqrt{\smash[b]2}/2, -\sqrt{\smash[b]2}/2)$

Now we have orthonormal basis

$$\{ (1/3, 2/3, 2/3), (2*\sqrt{\smash[b]2}/3, -\sqrt{\smash[b]2}/6, -\sqrt{\smash[b]2}/6) ,(0, \sqrt{\smash[b]2}/2, -\sqrt{\smash[b]2}/2) \}$$

and the only one thing we need to do is to define the matrix with rows being the vectors from that basis:

$\begin{pmatrix} {1 \over 3} & {2 \over 3} & {2 \over 3} \\ {2* \sqrt{\smash[b]2}\over 3} & {-\sqrt{\smash[b]2} \over 6} & {-\sqrt{\smash[b]2} \over 6}\\ 0 & {\sqrt{\smash[b]2} \over 2} & {-\sqrt{\smash[b]2} \over 2} \end{pmatrix}$