Normalization & value of Eigenvectors

  • Thread starter zak100
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  • #1
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Homework Statement


I have got the following matrix. I have found the eigen values but in some eq x, y & z terms are vanishing, so how to find the value of eigen vector??? Also why we have to do normalization??

A__=__[1__1__0]
______[1__1__0]
______[0__0__1]


Homework Equations


A-λI=0
Ax = -λIx

The Attempt at a Solution


A-λI=0
[1__1__0]_-_[λ__0__0]_____=0____________________
[1__1__0]___[0__λ__0]___________________________
[0__0__1]___[0__0__λ]___________________________
______________________________________________
[1-λ__1__0]=0__________________________________
[1___1-λ__0]____________________________________
[0___0____1-λ]__________________________________
_______________________________________________
1-λ|1-λ__0|____-1|1___0__| +0 =0__________________
___|0__1-λ|______|0___1-λ_|________________________
_______________________________________________
(1-λ)(1-λ)^2____-(1-λ) =0
Taking_(1-λ) common
(1-λ)[(1-λ)^2__-1]=0
First eigen value λ1 = 1
Now consider:
[(1-λ)^2_____-1]=0
1-2λ+λ^2-1=0
-2λ+λ^2=0
λ(λ-2)=0
λ2=0
& λ3=2
______________________________________________________
______________________________________________________
For λ1 = 1
Ax = λ*x*I
_____________________________________________________
[1__1__0]_______[x]___= 1 * [x]___________________________
[1__1__0]_______[y]_______[y]___________________________
[0__0__1]_______[z]_______[z]___________________________
______________________________________________________
x+y=x------eq(1)
x+y=y------eq(2)
z=z---------eq(3)
______________________________________________________
Above x vanishes in eq(1), y vanishes in eq(2) & z vanishes in eq(3).
What is the value of μ1?????
______________________________________________________
For λ2=0_______________________________________________
[1__1__0]_______[x]___= 0 * [x]____________________________
[1__1__0]_______[y]_______[y]____________________________
[0__0__1]_______[z]_______[z]____________________________
______________________________________________________
x+y=0----------eq(4)_______________________________________
x+y=0----------eq(5)_______________________________________
z=0-------------eq(6)_______________________________________
______________________________________________________
In my view it should be μ=[-1]_______________________________
_____________________[-1]______________________________
_____________________[0]_______________________________
but teacher has got different answer. In addition to this he has done normalization, please guide me what is the need for normalization_______________________________
______________________________________________________
For λ=2________________________________________________
_____________________________________________________
[1__1__0]_______[x]___=_2_*_ [x]___________________________
[1__1__0]_______[y]_______[y]___________________________
[0__0__1]_______[z]_______[z]___________________________
x+y=2x------eq(7)-----------------------------------------------------------------
x=-y----------------------------------------------------------------------------------
x+y=2y-------eq(8)----------------------------------------------------------------
y=x----------------------------------------------------------------------------------
z=2z----------eq(9)---------------------------------------------------------------
Therefor, μ=[-1]_________________________________________
__________[1]_________________________________________
__________[0]___________________________________________
Why we have to normalize the vector?
Somebody please guide me.
Zulfi.
 

Answers and Replies

  • #2
14,413
11,726
Above x vanishes in eq(1), y vanishes in eq(2) & z vanishes in eq(3).
No. It is correct that you get ##x=y=0##, but ##z=z## means that it is always true. So any value will do, i.e. ##(0,0,1)## is your eigenvector. It spans the eigenspace to the eigenvalue ##\lambda_1=1##, which means all vectors in ##\mathbb{R}\cdot (0,0,1)## are eigenvectors, which corresponds to the free value of ##z##. You don't have to choose ##z=1##, it is simply a matter of convenience.
x+y=0----------eq(4)_______________________________________
x+y=0----------eq(5)_______________________________________
z=0-------------eq(6)_______________________________________
______________________________________________________
In my view it should be μ=[-1]_______________________________
_____________________[-1]______________________________
_____________________[0]_______________________________
##x+y=0## means ##x=-y##. You have chosen ##x=y## which is wrong, and so is your eigenvector.
For λ=2________________________________________________
_____________________________________________________
[1__1__0]_______[x]___=_2_*_ [x]___________________________
[1__1__0]_______[y]_______[y]___________________________
[0__0__1]_______[z]_______[z]___________________________
x+y=2x------eq(7)-----------------------------------------------------------------
x=-y----------------------------------------------------------------------------------
x+y=2y-------eq(8)----------------------------------------------------------------
y=x----------------------------------------------------------------------------------
z=2z----------eq(9)---------------------------------------------------------------
Same problem. We have ##x=y## in both equations, so your ##x=-y## is wrong, and so is your eigenvector.
Why we have to normalize the vector?
See my explanation above. "Have to" is the wrong verb. Maybe you should try to prove, that if ##A\cdot \vec{x}=\lambda \cdot \vec{x}## then all ##\mu \cdot \vec{x}## are also eigenvectors to the eigenvalue ##\lambda##. To choose ##1's## as components makes computations easier.

Btw, this isn't a normalization, at least not according to the usual Euclidean norm. A normalized vector ##(1,1,0)## would be ##\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0 \right)\,.##
 
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  • #3
462
11
Hi,
Thanks for your reply.
λ1=1, eigen vector is:___________________________________________________
[0]__________________________________________________________________
[0]__________________________________________________________________
[1]___________________________________________________________________
because equations are:
x+y=x------eq(1) (No x-component, so x is 0
x+y=y------eq(2) (No y-component, so y is 0)
z=z---------eq(3) (Okay z=1)
_____________________________________________
I got it.
For λ2=0, I got:
For λ2=0_______________________________________________
[1__1__0]_______[x]___= 0 * [x]____________________________
[1__1__0]_______[y]_______[y]____________________________
[0__0__1]_______[z]_______[z]____________________________
______________________________________________________
x+y=0---------eq(4)_______________________________________
x+y=0----------eq(5)_______________________________________
z=0-------------eq(6)________
I cant understand what is the Eigen Vector in this case. Please guide me.

Zulfi.

Reference https://www.physicsforums.com/threads/normalization-value-of-eigenvectors.955610/

Reference https://www.physicsforums.com/threads/normalization-value-of-eigenvectors.955610/
 
  • #4
14,413
11,726
We have
$$
\begin{bmatrix}1&1&0\\1&1&0\\0&0&1\end{bmatrix}\cdot \begin{bmatrix}x\\y\\z\end{bmatrix}=0\cdot \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}
$$
and so ##x+y=0## and ##z=0##. So ##\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}x\\-x\\0\end{bmatrix}=x\cdot \begin{bmatrix}1\\-1\\0\end{bmatrix}##. If you did my exercise, you know that all multiples of an eigenvector is again an eigenvector to the same eigenvalue. Therefore any, say ##x-##multiple of ##(1,-1,0)^\tau## are all eigenvectors, especially the ##1-##fold of it (##x=1##).
 
  • #5
462
11
Hi,
Thanks for your response.
We have:
x+y=0 eq(4). So x is +ve in the eq so we would have +1.
But in eq(5)
x + y = 0
& we have to find y:
y = -x., so y =-1

& in last case, we have to find z, & z=0, so we have 0 in the 3rd entry of vector.
If in the 3rd eq, we were having:
x+z=0
So will z be equal to -1 in this case?
& if y+z = 0,
then again will z be zero?

Please guide me.
Zulfi.
 
  • #6
14,413
11,726
I'm not sure I understand you. Equations (4) and (5) are redundant, as they both yield ##x=-y##, and equation (6) is ##z=0##.
Since we wanted to calculate the vector ##\vec{x}=(x,y,z)## for which ##A\vec{x}=0\cdot \vec{x}=\vec{0}##, we get ##(x,y,z)=(x,-x,0)## which are all possible eigenvectors for the eigenvalue zero.
 
  • #7
462
11
Hi,
Thanks for your response. I cant understand how you got (x, -x, 0) from the equations?

Zulfi.
 
  • #8
14,413
11,726
Hi,
Thanks for your response. I cant understand how you got (x, -x, 0) from the equations?

Zulfi.
We have ##x+y=0##. This is the same as ##x+y-x=0-x## which is ##y=-x##. The last equation (6) directly says ##z=0##. So ##x## cannot be determined, or ##y## if you like it better. There is one free variable, I chose ##x##. And so ##(x,y,z)=(x,-x,0)\,.##
 
  • #9
462
11
Hi,
Thanks for your reply. So from eq:
x+y=0
we can have either x =-1 or y=-1. Is it possible to have (-1, 1, 0)

And once we get x =-1 we would put it into 2nd eq giving:
y=1.
And no problem with z.

Thanks.

Zulfi.
 
  • #10
14,413
11,726
Hi,
Thanks for your reply. So from eq:
x+y=0
we can have either x =-1 or y=-1. Is it possible to have (-1, 1, 0)

And once we get x =-1 we would put it into 2nd eq giving:
y=1.
And no problem with z.

Thanks.

Zulfi.
No, we do not get a certain value for ##x## or ##y## from the equations.

Let's change the notation, since you seem to confuse variables and values. We started with ##A\cdot \begin{bmatrix}u\\v\\w\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}## which is an equation full of numbers. Only thing is, we don't know the numbers ##u,v,w## yet. So we try to calculate them. This led us to the equations ##u+v=0## and ##w=0##. More cannot be said. So ##:=\begin{bmatrix}u\\v\\w\end{bmatrix}=\begin{bmatrix}u\\-u\\0\end{bmatrix}=u\cdot \begin{bmatrix}1\\-1\\0\end{bmatrix}##. But ##u## remains arbitrary, because we don't have another equation to fix it. All possible vectors ##u\cdot \begin{bmatrix}1\\-1\\0\end{bmatrix}## are eigenvectors, which means you can choose for ##u## whatever you like: with ##u=\pi## we get the eigenvector ##\begin{bmatrix}\pi \\ -\pi \\ 0 \end{bmatrix}##, for ##u=1## we get ##\begin{bmatrix}1\\-1\\0\end{bmatrix}##, for ##u=-1## we have ##\begin{bmatrix}-1\\1\\0\end{bmatrix}## and for ##u=0## we get ##\begin{bmatrix}0\\0\\0\end{bmatrix}## which is always an eigenvector to any eigenvalue. That's why it is usually not meant by the word eigenvector, nevertheless, it is one.

And I do not want to see the same sign on ##x## and ##y## again! How could their sum ever be zero, if they were both negative???
 
  • #11
462
11
Hi,
Thanks for your time. I highly appreciate yourself removing my confusion.

<And I do not want to see the same sign on x and y again!>
I have used the word either. Form the eq:
x+y =0
its clear that either x=-1 & y=1
or x= & y=-1.
which in short I said that either x =-1 or y=-1.

Also
<
we do not get a certain value for x or y from the equations. >

So why its necessary to have x= 1 & y=-1. It could also be possible that x=-1 & y=1.

Zulfi.
Reference https://www.physicsforums.com/threads/normalization-value-of-eigenvectors.955610/
Reference https://www.physicsforums.com/threads/normalization-value-of-eigenvectors.955610/
Reference https://www.physicsforums.com/threads/normalization-value-of-eigenvectors.955610/

Zulfi.
 
  • #12
14,413
11,726
So why its necessary to have x= 1 & y=-1. It could also be possible that x=-1 & y=1.
Yes, and many more (cp. my examples). ##x=0## is the only value which cannot be taken, as the zero vector is an eigenvector for all eigenvalues, and even for values, which are not eigenvalues, because ##A\cdot 0 = \mu \cdot 0## is always true, for any ##\mu##. So we need a value different form zero. ##x=1## is as good as any other. However, ##x=1## is far more convenient than ##x=15,613,687,516,498,984,816,098,780,098,732,198,075,519,158## would be. If the eigenvalue was ##\frac{1}{2}## it might be convenient to choose ##x=2##. It depends on what you want to do with the eigenvector. ##x=1## is simply the easiest one.
 

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