# Inner product with perpendicular unit vectors

• Mar 17th 2010, 06:36 PM
uberbandgeek6
Inner product with perpendicular unit vectors
Note: I'm using (A^T) for the transpose of A.

Problem: Find (A^T)A (the inner product of A with itself) if the columns of A are unit vectors, all mutually perpendicular.

My work: if the columns of A are a(1), a(2),..., a(n) then (A^T)A = ||a(1)||^2 + ||a(2)||^2 + ... + ||a(n)||^2 = 1+1+...+1 = n, so the answer is the number of columns (n) in A. Did I do this correctly?
• Mar 18th 2010, 03:45 AM
tonio
Quote:

Originally Posted by uberbandgeek6
Note: I'm using (A^T) for the transpose of A.

Problem: Find (A^T)A (the inner product of A with itself) if the columns of A are unit vectors, all mutually perpendicular.

My work: if the columns of A are a(1), a(2),..., a(n) then (A^T)A = ||a(1)||^2 + ||a(2)||^2 + ... + ||a(n)||^2 = 1+1+...+1 = n, so the answer is the number of columns (n) in A. Did I do this correctly?

Your line of thought is correct, but what do you mean by "the answer is the number of ..."?? The answer is a matrix and you haven't yet said which one \$\displaystyle A^TA\$ is !

Tonio
• Mar 18th 2010, 09:54 AM
uberbandgeek6
Oh, you're right, I guess it should be a matrix, but I don't understand how to get one. I just have a bunch of scalars now, right? How do I make a matrix out of that? Thank you for the catch btw!
• Mar 18th 2010, 01:35 PM
HallsofIvy
When taking a product of two matrices, AB, you can think of each element of the product matrix as a "dot product"- the "ij" element is the dot product of the ith row of A (as a vector) with the jth column of B (as a vector). Since these vectors are all of unit length and perpendicular, the dot product of each one with with itself is 1 and with any other is 0. Therefore each row of the product will consist of one "1" and the rest "0"s.