How to determine #a# and #b# so that the real matrix #((a,1,b),(b,a,1),(1,b,a))# is orthogonal matrix ?

2 Answers
Apr 9, 2017

#a=b=0#

Explanation:

A matrix #M# is orthogonal if and only if:

#MM^T = M^TM = I#

Evaluating this directly with the given form of matrix:

#((a,1,b),(b,a,1),(1,b,a))((a,1,b),(b,a,1),(1,b,a))^T#

#= ((a,1,b),(b,a,1),(1,b,a))((a,b,1),(1,a,b),(b,1,a))#

#=((a^2+b^2+1,ab+a+b,ab+a+b),(ab+a+b,a^2+b^2+1,ab+a+b),(ab+a+b,ab+a+b,a^2+b^2+1))#

In order for this to be the identity matrix, we require:

#a^2+b^2+1 = 1#

So if #a#, #b# are real then they must both be #0#.

Then we find:

#ab+a+b=0#

satisfying the requirement that the off diagonal elements be #0#.

Apr 9, 2017

See below

Explanation:

Also, an orthogonal matrix has columns and rows that are orthogonal unit vectors.

For the first column vector:

#abs(mathbfv_1) = sqrt (a^2 + b^2 + 1^2) = 1 implies a = b = 0 " if " a,b in RR#.