Eigenvalues and eigenvectors of a matrix

In summary, we are discussing eigenvectors and eigenvalues in a matrix, specifically in Z mod 7 and Z mod 11. We have found the characteristic polynomial and eigenvalues for both cases. To find the eigenvectors related to a specific eigenvalue, we need to solve a system of equations. In Z mod 7, we found the eigenvectors to be (4,1), (1,2), etc. In Z mod 11, we are struggling to find the correct solution, which is (1,8). We have realized that y=4x is a common mistake and the correct solution is (1,4).
  • #1
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Hello i have this matrix [itex]\in Z [/itex] mod [itex] 7[/itex],
M = \begin{pmatrix} 0&6\\ 5&0 \end{pmatrix}
always modulo [itex]7[/itex] in [itex]Z[/itex].
I found characteristic polynomial [itex]x^2+5[/itex].
Eigenvalues are [itex]\lambda = 3, \lambda' = 4[/itex]
Eigenvectors related to [itex]\lambda = 3 [/itex] are the non-zero solution of the system:
[itex]4x +6y = 0,[/itex]
[itex]5x+4y = 0 [/itex]
I get:
[itex]4x = y,[/itex]
[itex]6y[/itex]
I don't know if it is correct, but how can i find the eigenvectors?
 
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  • #2
Yes, y=4x is correct. Now you just need to find a non-zero vector (0,0) that satisfies this. So, we can take y=1 for example. In that case (x,y)=(4,1) satisfies the equation. So (4,1) is an eigenvector.
We could also take y=2, then y=(1,2) was an eigenvector. etc.
 
  • #3
now i have a problem to solve this:
in [itex]Z[/itex] mod [itex]11[/itex]

M = \begin{pmatrix} 0&2\\ 7&0 \end{pmatrix}
The characteristic polinomyal is : [itex]x^2+8[/itex],
eigenvalues [itex]\lambda=5, \lambda'=6[/itex],
eigenvectors related to [itex]\lambda=5[/itex] are the non-zero solution of the system:
[itex]6x+2y=0[/itex]
[itex]7x+6y=0[/itex]
I don't know how to solve this system because i took a look at the solution of the exercise and it is:
[itex]V={(y, 8y)}[/itex]
I don't know how to get this solution.
Oh i get now I'm wrong to write the first equation.:redface:
 
Last edited:
  • #4
micromass said:
Yes, y=4x is correct. Now you just need to find a non-zero vector (0,0) that satisfies this. So, we can take y=1 for example. In that case (x,y)=(4,1) satisfies the equation. So (4,1) is an eigenvector.
We could also take y=2, then y=(1,2) was an eigenvector. etc.
You have this backwards. If y= 4x, then the eigenvector is (1, 4), not (4, 1).
 
  • #5


Hello,

Thank you for sharing your work on finding the eigenvalues and eigenvectors of the given matrix. Your approach is correct, but there are a few steps missing that will help you find the eigenvectors.

To find the eigenvectors, we need to solve the system of equations you have written for the corresponding eigenvalue. In this case, we have the system:

4x + 6y = 0
5x + 4y = 0

To solve this system, we can use any method we prefer, such as substitution or elimination. I will use substitution in this response.

First, we can rearrange the first equation to get y in terms of x:

y = -4/6 x

Now, we can substitute this value of y into the second equation:

5x + 4(-4/6x) = 0
5x - 8/3x = 0
(5 - 8/3)x = 0
x = 0 or x = 0

Since we cannot have zero as an eigenvector, we can ignore the first solution and focus on x = 0. Now, we can plug this value of x into the first equation to find the corresponding value of y:

4(0) + 6y = 0
y = 0

Therefore, the eigenvector corresponding to the eigenvalue \lambda = 3 is [0, 0]. You can follow a similar approach to find the eigenvector corresponding to the eigenvalue \lambda' = 4.

I hope this helps clarify the process of finding eigenvectors. Keep up the good work!
 

1. What are eigenvalues and eigenvectors?

Eigenvalues and eigenvectors are mathematical concepts used in linear algebra to describe the behavior of a matrix. Eigenvalues are the scalars that represent the scaling factor of the eigenvectors, which are the non-zero vectors that remain in the same direction after being multiplied by the matrix.

2. How are eigenvalues and eigenvectors calculated?

Eigenvalues and eigenvectors are calculated by solving the characteristic equation of a matrix, which is obtained by subtracting the identity matrix from the original matrix and finding its determinant. The eigenvectors are then found by solving a system of linear equations using the eigenvalues as coefficients.

3. What are the applications of eigenvalues and eigenvectors?

Eigenvalues and eigenvectors have various applications in fields such as physics, engineering, and data analysis. They are used to describe the behavior of linear systems, find dominant patterns in data, and solve differential equations.

4. Can a matrix have more than one set of eigenvalues and eigenvectors?

Yes, a matrix can have multiple sets of eigenvalues and eigenvectors. This is because for a matrix to have eigenvalues and eigenvectors, it must be square and diagonalizable. Diagonalizable matrices can have multiple eigenvectors for the same eigenvalue, resulting in different sets of eigenvalues and eigenvectors.

5. How are eigenvalues and eigenvectors related to the diagonalization of a matrix?

Eigenvalues and eigenvectors play a crucial role in the diagonalization of a matrix, which is the process of finding a similar matrix that is diagonal. This is done by using the eigenvectors as the columns of the new matrix and the eigenvalues as the diagonal entries. Diagonalization simplifies the calculation of powers and inverses of a matrix.

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