Gram-Schmidt process to find last row of matrix

**sfspitfire23** · October 20th 2012, 06:58 PM

Hi guys,

Use the Gram-Schmidt process to find last row of the following orthogonal matrix where the first two rows are:

[sqrt(3)/2, -1/sqrt(8), 1/sqrt(8)]

and

[0, 1/sqrt(3), sqrt(2)/sqrt(3)]

(third row is [a,b,c]).

Now, because the matrix is orthogonal, all of the rows are perpendicular and have length 1. To find the third row, call it q3, using Gram-Schmidt, we would do this-

q3 = [a,b,c] - proj q1 ([a,b,c]) - proj q2 ([a,b,c])

where q1 = [sqrt(3)/2, -1/sqrt(8), 1/sqrt(8)] and q2 = [0, 1/sqrt(3), sqrt(2)/sqrt(3)].

When I try and brute force this thing and solve, I get crazy fractions etc. I must be missing something...

**MaxJasper** · October 20th 2012, 09:38 PM

$p = \left(\begin{array}{ccc} \frac{\sqrt{3}}{2} & -\frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} \\ 0 & \frac{1}{\sqrt{3}} & \sqrt{\frac{2}{3}} \\ a & b & c\end{array}\right)$

p=orthogonal which means:

$p^{\mathsf{T}}.p=I$

Orthogonalize each column, and solve above identity to find a,b,c making p:

$p = \left(\begin{array}{ccc} \frac{\sqrt{3}}{2} & -\frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} \\ 0 & \frac{1}{\sqrt{3}} & \sqrt{\frac{2}{3}} \\ \frac{3}{2} \sqrt{\frac{1}{119} \left(3+8 \sqrt{2}\right)} & \frac{16 \sqrt{357 \left(3+8 \sqrt{2}\right)}-3 \sqrt{714 \left(3+8 \sqrt{2}\right)}}{1428} & \frac{-16 \sqrt{357 \left(3+8 \sqrt{2}\right)}+3 \sqrt{714 \left(3+8 \sqrt{2}\right)}}{1428}\end{array}\right)$

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