Eigenvalues and Eigenvectors question

In summary, the conversation discusses finding the eigenvectors for a given 3x3 matrix with eigenvalues of a, 0, and 1. The process involves setting one of the components in the eigenvector to 1 and solving for the other two components using two equations obtained from the matrix. However, in some cases, setting a component to 1 may not lead to a solution, in which case another component can be chosen. It is important to note that the matrix used in the process must be correct and care must be taken when setting components equal to 1, as this may affect the solution.
  • #1
Dank2
213
4
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
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
andrewkirk said:
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
Dank2 said:
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
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
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
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
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
i get in my second equation bx = a, but i can't divide by b since b can be equal to 0. only a is not equal to 1 or 0. b in R .
 
  • #10
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
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
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
andrewkirk said:
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 therefore 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
@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
andrewkirk said:
@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 as well and there will be a solution. same problem as in post # 9
 
  • #16
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 got to put them in P such that
P'AP = Diagonal.
 
  • #17
Dank2 said:
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
$$y=\frac ba x$$
 

1. What are eigenvalues and eigenvectors?

Eigenvalues and eigenvectors are mathematical concepts used to understand the behavior of linear transformations on vector spaces. Eigenvalues represent the scalar values that a linear transformation scales the corresponding eigenvectors by, while eigenvectors represent the non-zero vectors that remain in the same direction after a linear transformation.

2. How are eigenvalues and eigenvectors calculated?

Eigenvalues and eigenvectors can be calculated by solving the characteristic equation of a given linear transformation. The characteristic equation is obtained by setting the determinant of the transformation's matrix equal to zero and solving for the corresponding eigenvalues. The eigenvectors can then be found by plugging the eigenvalues back into the original equation and solving for the non-zero vectors.

3. What are the applications of eigenvalues and eigenvectors?

Eigenvalues and eigenvectors have various applications in different fields such as physics, engineering, economics, and computer science. They are used to understand the stability of dynamic systems, analyze data in machine learning algorithms, and solve differential equations. They are also used in image processing, signal processing, and quantum mechanics.

4. How do eigenvalues and eigenvectors relate to diagonalization?

Eigenvalues and eigenvectors are essential in diagonalization, which is the process of finding a diagonal matrix that is similar to a given square matrix. Diagonalization is useful in simplifying the computations of matrix operations and solving systems of linear equations. Eigenvalues are the diagonal entries of the diagonal matrix, and eigenvectors form the columns of the matrix that diagonalizes the original matrix.

5. Can there be multiple eigenvalues and eigenvectors for a given linear transformation?

Yes, it is possible for a linear transformation to have multiple eigenvalues and corresponding eigenvectors. This occurs when the matrix of the linear transformation has repeated eigenvalues, or when the matrix is not diagonalizable. In these cases, there can be multiple eigenvectors associated with the same eigenvalue, and they form a subspace called the eigenspace.

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