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HELP: orthogonal sets & orthogonal matrices HW problem

  • Thread starter xcvxcvvc
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Given a = ( 1, -2, 1), b= ( 0, 1, 2), and c = (-5, -2, 1) determine if {a, b, c} is an orthogonal set. Show support for your answer.

I know that if the dot product of every combination equals zero, the set is orthogonal. No problems here. I do that, and they all equal zero.
a dot b:
1 * 0 + -2 * 1 + 1 * 2 = 0
a dot c:
1 * -5 + -2 * -2 + 1 * 1 = 0
b dot c:
0 * -5 + 1 * -2 + 2 * 1 = 0

The problem comes with another theorem I learned to determine if a set is orthogonal. I thought if i made a square matrix out of orthogonal sets, the matrix is also orthogonal. Thus, the matrix's transpose equals its inverse and its transpose times the original equals an identity matrix.

So I check what I've already concluded this time with the matrix method:

[ 1, -2, 1]T [1, -2, 1] --------------------[1,0,0]
[ 0, 1, 2] * [ 0, 1, 2] which doesn't equal [0,1,0]
[-5, -2, 1] [-5, -2, 1] -------------------- [0,0,1]

I tried making the vectors columns and rows(from my understanding, both should work), and there is an example in my notes that uses this method. The example works. this does not. It's really frustrating.

I'm using a calculator to do the matrix math.
 

HallsofIvy

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Given a = ( 1, -2, 1), b= ( 0, 1, 2), and c = (-5, -2, 1) determine if {a, b, c} is an orthogonal set. Show support for your answer.

I know that if the dot product of every combination equals zero, the set is orthogonal. No problems here. I do that, and they all equal zero.
a dot b:
1 * 0 + -2 * 1 + 1 * 2 = 0
a dot c:
1 * -5 + -2 * -2 + 1 * 1 = 0
b dot c:
0 * -5 + 1 * -2 + 2 * 1 = 0

The problem comes with another theorem I learned to determine if a set is orthogonal. I thought if i made a square matrix out of orthogonal sets, the matrix is also orthogonal.
No! If a set of vectors is orthonormal, then a matrix having those vectors as columns or rows is orthogonal. The vectors you give are orthogonal but not "normal"- they do not have length 1. If you were to divide each of those vectors by its length, then you would have an orthonormal set and they would form an orthogonal matrix.

Thus, the matrix's transpose equals its inverse and its transpose times the original equals an identity matrix.

So I check what I've already concluded this time with the matrix method:

[ 1, -2, 1]T [1, -2, 1] --------------------[1,0,0]
[ 0, 1, 2] * [ 0, 1, 2] which doesn't equal [0,1,0]
[-5, -2, 1] [-5, -2, 1] -------------------- [0,0,1]

I tried making the vectors columns and rows(from my understanding, both should work), and there is an example in my notes that uses this method. The example works. this does not. It's really frustrating.

I'm using a calculator to do the matrix math.
 
No! If a set of vectors is orthonormal, then a matrix having those vectors as columns or rows is orthogonal. The vectors you give are orthogonal but not "normal"- they do not have length 1. If you were to divide each of those vectors by its length, then you would have an orthonormal set and they would form an orthogonal matrix.
ahh, thanks. Is there any pattern formed when a matrix's rows or columns are orthogonal vectors? I know when I arrange those orthogonal vectors into columns, the transpose * the original equals a diagonal matrix. Yet, arranging the orthogonal vectors into rows gives me a mess of a matrix no where near diagonal.

Also, the orthonormal vector identity matrix orthogonal matrix thing only works for square matrices. right?
 

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