Question

If a square matrix A s.t. A^tA = I, then detA=1 or -1

True, but why?

Answer 1

Isnt A^t A = I always true?

Answer 2

Nope, \(A^TA = I\) is not always true, suppose in the simplest case A is a matrix full of 1's. then the transpose is also another matrix full of ones. Multiplying them together won't get you an identity matrix since there's no subtraction happening to get any zeroes!

Answer 3

detA = 1/det(A^t)
detA = 1/(detA)

Answer 4

oh

detA detA = 1
(detA)^2 = 1

det A = 1 or -1

Answer 5

Yeah, I guess I really should have mentioned it but you figured it out, that det(A^t)=det(A).

Answer 6

While you're on the subject, \(A^TA\) is always a symmetric matrix since it's equal to its own transpose which is kinda cool.

\((A^TA)^T=A^TA^{TT}=A^TA\)

Answer 7

Not only that, I just realized that if it also obeys the rule you said above, that is \(A^TA=I\) then we have a separate symmetric matrix that's idempotent, as in when you multiply it by itself you get the matrix back. Here I'll show:

\(B=AA^T\)
\(B=B^T\) Symmetric, check similar to how I did in the last post
\(BB=B\) Idempotent, I'll show below:

\(BB=AA^TAA^T =A I A^T = AA^T = B\)

This is used in quantum mechanics when using hermitian operators (complex version of symmetric matrices) and I thought it'd be nice to at least say that there's a compelling reason to appreciate these haha.