# Orthonormal vectors

1. Dec 2, 2007

### oswaler

1. The problem statement, all variables and given/known data

for 2 matrices, A and B

A=
a 0 0
0 -a 0
0 0 -a

B=
b 0 0
0 0 -ib
0 ib 0

Find three orthonormal vectors which are simultaneous eigenvectors of A and B.

2. Relevant equations

3. The attempt at a solution

I am very new to matrix math. The first thing that confused me was the use of the term "simultaneous eigenvectors." I looked through a bunch of books without finding it, so I assume the question is asking in general for 3 vectors that are also eigenvectors of the matrices. I understand the term orthonormal (both orthogonal and normalized). I'm kind of stuck on where to start. I tried getting eigenvalues for the matrices but the answers didn't seem to make sense.

Any help would be greatly appreciated.

-Eric
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 2, 2007

### EnumaElish

What did you try? What did you get? Why did the answers not make sense?

3. Dec 2, 2007

### oswaler

I don't know how to put the lambda symbol in here. Let #=lambda

I took AX=#X

so
a-# 0 0
0 -a-# 0
0 0 -a-#

(a-#)(-a-#)^2=0

so #=-a,a

But there should be 3 eigenvalues. If I use these values to get the eigenvectors, I get both vectors = (0 0 0)

When I do the same for the B matrix I get the same result.

Like I say, I'm very new to this. We got this stuff sort of thrown at us at the last minute and I can't find a good similar example in the textbooks I have.

Thanks - Eric

4. Dec 3, 2007

### HallsofIvy

Staff Emeritus
Yes, those are the eigenvalues. There do NOT have to be 3 distinct eigenvalues- here -a is a "double root" and you can expect (since the matrix is symmetric) to find two independent eigenvectors corresponding to $\lambda= -a$. Unfortunately, you have left out the crucial part: you say, " If I use these values to get the eigenvectors, I get both vectors = (0 0 0)". The whole point of "eigenvalue" is that the equation Av= $\lambda$v has non-trivial solutions. (0 0 0) is the trivial solution. HOW did you attempt to find the eigenvectors? If I multiply out
$$\left[\begin{array}{ccc} a & 0 & 0 \\ 0 & -a & 0 \\ 0 & 0 & -a\end{array}\right]\left[\begin{array}{c} x \\ y \\ z\end{array}\right]$$
I get [ax -ay -az]. Setting that equal to av= [ax ay az], I get three equations: ax= ax, -ay= ay, and -az= az. Obviously, -ay= ay has only y= 0 as solution and -az= az has only z= 0 as solution, but ax= ax has any x as solution: an eigenvalue is [x 0 0] for any x. Taking $\lambda= -a$ gives [0 y 0] and [0 0 z] as eigenvalues for any y or z.

First, you have to understand what "eigenvalues" are. If your statement "If I use these values to get the eigenvectors, I get both vectors = (0 0 0)" were true, then you would be saying they are not eigenvalues!

One thing you clearly need to understand, that has nothing to do with linear algebra, is that the equation ax= ax has every number x as solution!

Last edited: Dec 3, 2007
5. Dec 3, 2007

### oswaler

Thank you. So I see I now have 3 eigenvectors for A: (1 0 0) (0 1 0) (0 0 1), which are obviously orthonormal.

When I get the eigenvalues for B I have b, -b, -b

When I set Av=#v for #=b I get 3 equations:
bx=bx
iby=by
-ibz=bz

so (x 0 0)

but for #=-b:
bx=-bx (so x=0)
-ibz=-by
iby=-bz

so y=iz and iy=-z (multiplying the 2nd equation by i on both sides again gives y=iz)
which seems like it would give a vector (0 i 1).

So I'm obviously missing something since the eigenvectors for both matrices should be the same.