Eigenvalues and Eigenvectors question

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  • #1
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Let's say i have the 3x3 matrix
Code:
a  0  0
b  0  0
1  2  1
it's eigenvalues are e1 =a, e2 = 0, e3 = 1.
now if a = / = 0, 1
i have 3 distinct eigenvalues and the matrix is surely can be Diagonalizable.
so if i try to solve for the eigenvector for the eigenvalue e1 =a:
Code:
0  0  0
b  -a  0
1  2  1-a
after doing elementary operations i get:
Code:
1        2      1-a
0   -a-2b    -b(1-a)
0      0        0
now if -a-2b =0, then -b(1-a) =/=0, and vice versa. how do i find the eigenvector for a=/=0,1, if i don't know which one is equal to zero A22 or A23 ?
 
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  • #2
andrewkirk
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Your second code panel gives you two equations in the three unknowns (x,y,z) that are the components of the sought eigenvector. Set any of the three components to 1 to remove one unknown and then solve the system of two equations in two unknowns to find what the corresponding values of the other two components must be. Eigenvectors with eigenvalue a will be any multiple of that vector.
 
  • #3
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Your second code panel gives you two equations in the three unknowns (x,y,z) that are the components of the sought eigenvector. Set any of the three components to 1 to remove one unknown and then solve the system of two equations in two unknowns to find what the corresponding values of the other two components must be. Eigenvectors with eigenvalue a will be any multiple of that vector.
You mean should i put values to a and b and then solve numerically? because if i just multiply by (x,1,z) i get that a = 0:
1 2 1-a
0 -a-2b -b(1-a)
0 0 0
i got the eigenvector v= ((3-a)/(a-1) , 1 , t) t in R.
but it doesn't worth the matrix * v= a * v
 
  • #4
andrewkirk
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You mean should i put values to a and b and then solve numerically? because if i just multiply by (x,1,z) i get that a = 0:
1 2 1-a
0 -a-2b -b(1-a)
0 0 0
You seem to be using the matrix in your third code panel. My suggestion was to use the matrix in the second code panel, which I am confident is correct. I have not checked the matrix in the third panel and do not know whether it is correct.

Also, it sometimes happens that setting a particular component to 1 in the candidate eigenvector does not lead to a solution. In that case, one just chooses a different component to set to 1. There will be at least one component that can generate a solution by setting it to 1. But in this case I'm fairly confident there will be no problem with using the second component, provided the correct matrix is used.
 
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  • #5
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If i just solve i got
0 = a*x
bx-ay = y*a
x+2y+z-az = z*a

from first eq i get, x = 0, since a=/=0, 1
then y = 0, since -ay = ay and a=/=0, 1
any more suggestions how can i continue ?
 
  • #6
andrewkirk
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I'm afraid I don't understand what calculations you are doing. Based on the matrix in the second code panel, the multiplication to find the eigenvector is:

$$\begin{pmatrix}
0 & 0 & 0 \\
b & -a & 0 \\
1 & 2 & (1-a)
\end{pmatrix}
\begin{pmatrix}x\\1\\z\end{pmatrix}=
\begin{pmatrix}0\\0\\0\end{pmatrix}
$$
That does not generate the same equations as you have written.

Ah, I just noticed. I think the problem might be that you are setting the result of the multiplication equal to ##\begin{pmatrix}x\\1\\z\end{pmatrix}##, whereas actually it should be ##\begin{pmatrix}0\\0\\0\end{pmatrix}##, as the former vector has already been deducted from both sides to obtain the matrix in the second panel. Also, remember to set y equal to 1 so that there are only two variables (x and z) to be found from the two equations.
 
  • #7
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ImageUploadedByPhysics Forums1466681241.772585.jpg

I think you thought i solve for a = 0 . I need to find eigenvalue a that is different from 0 or 1. I've uploaded a pic
do i must put numerical values in a and b in order to get the eigenvector that belongs to the eigenvalue a? i wanted to find the eigenvector in terms of a and b.
 
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  • #8
andrewkirk
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That equation you have drawn is incorrect. Correct versions are the one I posted in my previous post or, alternatively (and equivalently):

$$
\begin{pmatrix}
a & 0 & 0 \\
b & 0 & 0 \\
1 & 2 & 1
\end{pmatrix}
\begin{pmatrix}x\\1\\z\end{pmatrix}=
a\cdot\begin{pmatrix}x\\1\\z\end{pmatrix}
$$

You solve this for ##x## and ##z##, with the solutions being written as formulas that use ##a## and ##b##. You do not replace ##a## or ##b## by numerical values.
 
  • #9
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i get in my second equation bx = a, but i can't devide by b since b can be equal to 0. only a is not equal to 1 or 0. b in R .
 
  • #10
andrewkirk
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If you set b=0 and solve the equations you will deduce that a=0. Since that contradicts the premise that a is nonzero, you can conclude (by the contrapositive) that b must also be nonzero. Hence you can divide by b.
 
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  • #11
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but for any if a = 2, b = 0, the matrix is still diagonalizable, if i set b =/= 0 i exclude all the possiblities where b= 0 and a in R =/= 0,1.
so it's not the all of the solution.
 
  • #12
andrewkirk
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My mistake. If ##b=0## our conclusion is not that ##a=0## but rather that the eigenvector corresponding to eigenvalue ##a## cannot have 1 as the second component. So, as foreshadowed in post #4, let's instead set the third component equal to 1 and solve
$$
\begin{pmatrix}
a & 0 & 0 \\
b & 0& 0 \\
1 & 2 & 1
\end{pmatrix}
\begin{pmatrix}x\\y\\1\end{pmatrix}=
a \begin{pmatrix}x\\y\\1\end{pmatrix}
$$
 
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  • #13
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My mistake. If ##b=0## our conclusion is not that ##a=0## but rather that the eigenvector corresponding to eigenvalue ##a## cannot have 1 as the second component. So, as foreshadowed in post #4, let's instead set the third component equal to 1 and solve
$$
\begin{pmatrix}
a & 0 & 0 \\
b & 0& 0 \\
1 & 2 & 1
\end{pmatrix}
\begin{pmatrix}x\\y\\1\end{pmatrix}=
a \begin{pmatrix}x\\y\\1\end{pmatrix}
$$
It did work now, can you please explain me why i could set z to 1? because the eigenvector is subspace and therfore i can multiply the eigenvector by 1/x or 1/y or 1/z and it would still be eigenvector ? is that the reason? since A*v = a*v <==> A*1/x*v = 1/x*a*v ?
 
  • #14
andrewkirk
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@Dank2 Yes you can multiply an eigenvector by any nonzero scalar and it remains an eigenvector. So in searching for the eigenvector corresponding to a known eigenvalue, we can reduce the number of unknowns in our system of equations by setting a component of the candidate eigenvector to 1 as long as that component is not zero in the solution.
 
  • #15
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@Dank2 Yes you can multiply an eigenvector by any nonzero scalar and it remains an eigenvector. So in searching for the eigenvector corresponding to a known eigenvalue, we can reduce the number of unknowns in our system of equations by setting a component of the candidate eigenvector to 1 as long as that component is not zero in the solution.
One last thing that i have noticed.
In getting the solution :
x = (a^2-a)/(a+2b)
y = (ba-b)/(a+2b)
I have noticed that i devided by b in order to solve for x. but b is in R, it can be 0 aswell and there will be a solution. same problem as in post # 9
 
  • #16
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I think since i have a and b and in the matrix , i cannot find one eigenvector for both of them, since ill get different ones for different values, and i need to find a matrix that diagonalize A, and so i think i must put numeric values in a and b for that . , since i need only one eigenvector for the eigenvalue a =/= 0, 1. and two other eigenvectors for other eigenvalues i have. i gotta put them in P such that
P'AP = Diagonal.
 
  • #17
andrewkirk
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I have noticed that i devided by b in order to solve for x.
You shouldn't need to do that. From the second component equation you get
$$bx=ay$$
But we can divide by ##a## instead of ##b##, since we are told that ##a\neq 0##. That gives us
$$y=\frac ba x$$
 

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