This is driving me mad!
I have to find the eigenvectors of the matrix A=[1, 0 : 1, 2 ]
So first I find the eigenvalues by solving the det (A  kI) = 0
Where k = lamda , the eigenvalue and I is the identity matrix.
So I end up with [ 1k , 0 : 1 , 2k] which gives the characteristic polynomial as:
k^2 2k +2 = 0
So k=2 ,1
Then To find eigenvectors I solve (AkI)v = 0
Where v is the eigenvector.
So if I sub my eigenvalues back into the AkI matrix and then solve I SHOULD get my eigenvectors. BUT my matrix becomes:
for k=1
[0,0 : 1,1] [v1:v2] = [0:0]
Giving:
0x + 0y =0
which to me gives the eigenvector [1:1] not to sure why just thats what others have looked like but the actual answers are:
[0:1] and 1/sqrt(2) *[1:1]
HOW?

Carlo08
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 02122010 23:56

Clarity Incognito
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 03122010 00:10
(Original post by Carlo08)
This is driving me mad!
I have to find the eigenvectors of the matrix A=[1, 0 : 1, 2 ]
So first I find the eigenvalues by solving the det (A  kI) = 0
Where k = lamda , the eigenvalue and I is the identity matrix.
So I end up with [ 1k , 0 : 1 , 2k] which gives the characteristic polynomial as:
k^2 2k +2 = 0
So k=2 ,1
Then To find eigenvectors I solve (AkI)v = 0
Where v is the eigenvector.
So if I sub my eigenvalues back into the AkI matrix and then solve I SHOULD get my eigenvectors. BUT my matrix becomes:
for k=1
[0,0 : 1,1] [v1:v2] = [0:0]
Giving:
0x + 0y =0
which to me gives the eigenvector [1:1] not to sure why just thats what others have looked like but the actual answers are:
[0:1] and 1/sqrt(2) *[1:1]
HOW?
Eigenvectors that satisfy the equation are and where k is an arbitrary constant. Of course, is a trivial solution. 
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 03122010 00:29
(Original post by Clarity Incognito)
Eigenvectors that satisfy the equation are and where k is an arbitrary constant. Of course, is a trivial solution. 
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 03122010 00:36
(Original post by Carlo08)
I'm sorry I have no idea what you mean here.
Expanding and rearranging we have:
Av = kv where k is an eigenvalue and v is the eigenvector corresponding to the eigenvalue.
v = where we want to know what x and y are right?
We get the two equations:
x = 2x
x+2y=2y
Solve, you'll find that x must be 0 and y can be anything.
Do the same for
x = x
x+ 2y = y ( y=x)
If you chose x=800 then y=800 or
and are both normalised eigenvectors, that might be specified in the question.Last edited by Clarity Incognito; 03122010 at 00:40. 
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 03122010 00:55
(Original post by Clarity Incognito)
(AkI)v=0
Expanding and rearranging we have:
Av = kv where k is an eigenvalue and v is the eigenvector corresponding to the eigenvalue.
v = where we want to know what x and y are right?
We get the two equations:
x = 2x
x+2y=2y
Solve, you'll find that x must be 0 and y can be anything.
Do the same for
x = x
x+ 2y = y ( y=x)
If you chose x=800 then y=800 or
and are both normalised eigenvectors, that might be specified in the question. 
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 03122010 01:06
(Original post by Carlo08)
Did you get different eigenvalues than I did, or did you just pick two for example? Mine where 2 and 1 which for k=1 gave me the equations : 0x+0y = x and x+y=y
I might be taking your matrix A to be different though
A =
or
A = ? Are you multiplying matrix A with correctly? where are the two 0s coming from in the first equation?Last edited by Clarity Incognito; 03122010 at 01:08. 
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 03122010 01:49
(Original post by Clarity Incognito)
Yeah, I got k=2 and 1 as well
I might be taking your matrix A to be different though
A =
or
A = ? Are you multiplying matrix A with correctly? where are the two 0s coming from in the first equation?
and then when I put the value 1 back into the matrix as k I get
\begin{pmatrix} 11 & 0 \\ 1 & 21 \end{pmatrix}
giving:
\begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix}
This is correct right?Last edited by Carlo08; 03122010 at 01:53. 
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 03122010 02:23
(Original post by Carlo08)
and then when I put the value 1 back into the matrix as k I get
giving:
This is correct right?
Yeah, I see what you've done now, so the equations you have now are
giving x +y = 0 i.e. x = y. You can now choose any value for x or y and the corresponding y or x value. The easiest being 1 giving They have then normalised this vector.
For the eigenvalue k=2 we have giving x=0 and x=0 and no conditions for y i.e. y can be anything but x must be 0. And so we have The most straightforward being y=1 as y=0 gives the trivial zero vector. With y =1, this is already a normalised eigenvector.Last edited by Clarity Incognito; 03122010 at 02:35. 
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 03122010 03:14
(Original post by Clarity Incognito)
You've got to add latex] or tex] at the start of latex statements and when you're done, finish with /latex] or /tex] (all the latex commands start with "[")
Yeah, I see what you've done now, so the equations you have now are
giving x +y = 0 i.e. x = y. You can now choose any value for x or y and the corresponding y or x value. The easiest being 1 giving They have then normalised this vector.
For the eigenvalue k=2 we have giving x=0 and x=0 and no conditions for y i.e. y can be anything but x must be 0. And so we have The most straightforward being y=1 as y=0 gives the trivial zero vector. With y =1, this is already a normalised eigenvector.
Ah thank you, your explanation of the first answer makes perfect sense to me. So basically I want to simplify the equations I get into x=y and then just give a value to x and see what the corresponding y value is. How does this work if I have 2 equations, or will they both be multiples of eachother? ie x=y and 4x=4y kind of thing?
The answer given in my notes for the first one is, as you showed me [1:1] however its multiplied by 1/sqrt(2) why is this .... and for k=2 it has been given as [0:1] as you said, thank you 
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 03122010 03:33
(Original post by Carlo08)
Ah thank you, your explanation of the first answer makes perfect sense to me. So basically I want to simplify the equations I get into x=y and then just give a value to x and see what the corresponding y value is. How does this work if I have 2 equations, or will they both be multiples of eachother? ie x=y and 4x=4y kind of thing?
The answer given in my notes for the first one is, as you showed me [1:1] however its multiplied by 1/sqrt(2) why is this .... and for k=2 it has been given as [0:1] as you said, thank you
It's multiplied by 1/sqrt2 to 'normalise' the eigenvector. A normalised vector is a vector of unit magnitude. You normalise a vector by dividing each element by the magnitude of the vector, which is
So the eigenvector has a magnitude of Dividing each element of the eigenvector by this you getLast edited by Clarity Incognito; 03122010 at 03:34. 
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 03122010 12:35
(Original post by Clarity Incognito)
Yeah, if there are two equations then you just solve for x and y like you would for standard simultaneous equations, if one of the equations was x=y then yes the other could be a multiple of it.
It's multiplied by 1/sqrt2 to 'normalise' the eigenvector. A normalised vector is a vector of unit magnitude. You normalise a vector by dividing each element by the magnitude of the vector, which is
So the eigenvector has a magnitude of Dividing each element of the eigenvector by this you get
Thank you 
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 04122010 00:03
(Original post by Carlo08)
Thank you so so much for all that help! I missed the lecture on it and have been trying to learn it on my own but with your help I not understand it. When would I know to normalise the vector or not? It doesn't say in the question, would I just always do it?
Thank you
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