Eigenvector of A_n: Show & Find Eigenvalue

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In summary, Eigenvalues and eigenvectors are important tools in many areas of mathematics and are used to solve problems involving linear transformations and matrices. They can also be applied to determine the characteristics of quadratic functions and to predict the direction of stress in objects.
  • #1
FourierX
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Homework Statement



Directly show that the n x 1 vector [1 1 1 ...1]T is an eigenvector of An. What is its associated eigenvalues?

Homework Equations



N/A

The Attempt at a Solution



I started going over this topic since we did not cover it in class due to time constraints and I do not know much about it. Any additional link would be highly appreciated.
 
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  • #2
What is [itex]A_n[/itex]?
 
  • #3
Sorry, my bad. A_n is a matrix (It is pretty long and I could not write it in the forum)

A_n =
[1 . ...1]
[. . ....1]
[. . ....1] , ...
[. . ....1]
[1 1 ...1]
 
  • #4
Is that supposed to be an n x n matrix with all entries equal to 1?

Suppose v is an eigenvector of A_n, what can you say about A_nv?
 
  • #5
A_n is a n x n matrix with all elements 1.



A_nv = [tex]\lambda[/tex]v

where,
[tex]\lambda[/tex] = eigvenvalue
?
 
  • #6
Yes. So does A_n times [1 1 1 ...1]T satisfy that condition?
 
  • #7
A_n * [1 1 1 1...1]T gives [n n n n...n], doesn't it ?
 
  • #8
Actually it gives [n n n n...n]T :wink:

Is that a scalar multiple of [1 1 1 1...1]T? If so, what is the eigenvalue?
 
  • #9
haha, yeah right, i forgot ^T.

that is a scalar multiple of [1 1 1 1...1]^T, which means n is the associated eigenvalue. Is it that easy ?
 
  • #10
Yup :smile:

A_n [1 1 1 1...1]^T=n[1 1 1 1...1]^T, so [1 1 1 1...1]^T must be an eigenvector of A_n with eigenvalue n. Simple as that.
 
  • #11
Wow. Thanks a lot!

Could you also give me a slight hint on application of Eigenvalues and vectors?
 
  • #12
A "slight hint"? Eigenvalues and eigenvectors are used throughout mathematics. To take one simple example, "linear differential operators" are linear transformations on vector spaces of functions and solutions to linear differential equations are always based on eigenvalues and eigenvectors of the operators.

One method of determining what kinds of graphs (hyperbola, ellipse, or parabola, ...) quadratic functions give is to rewrite them as a matrix equation and find the eigenvalues (which give coefficients) and eigenvectors (which give the "principal" axes).

To find along which lines objects under stress are likely to crack first, write their "strain tensors" as matrices and find the eigenvalues and eigenvectors.
 

1. What is an eigenvector and how is it related to A_n?

An eigenvector is a vector that does not change direction when multiplied by a matrix. In other words, it is a special vector that remains in the same direction after a transformation. In the context of A_n, an eigenvector is a vector that, when multiplied by the matrix A_n, yields a scalar multiple of itself. This scalar multiple is known as the eigenvalue.

2. How can I find the eigenvector and eigenvalue of A_n?

To find the eigenvector and eigenvalue of A_n, you can follow these steps:

  1. Find the characteristic polynomial of A_n by taking the determinant of (A_n - λI), where λ is a variable.
  2. Solve the characteristic polynomial for its roots, which will be the eigenvalues of A_n.
  3. For each eigenvalue, solve the equation (A_n - λI)x = 0 to find the corresponding eigenvector.

3. Can A_n have more than one eigenvector for a given eigenvalue?

Yes, A_n can have multiple eigenvectors for a given eigenvalue. In fact, there can be an infinite number of eigenvectors for a single eigenvalue. This is because any scalar multiple of an eigenvector is also an eigenvector for the same eigenvalue.

4. How do I show that a given vector is an eigenvector of A_n?

To show that a vector v is an eigenvector of A_n, you can multiply v by A_n and see if the result is a scalar multiple of v. In other words, if Av = λv, where λ is a scalar, then v is an eigenvector of A_n with eigenvalue λ.

5. Can A_n have complex eigenvalues and eigenvectors?

Yes, A_n can have complex eigenvalues and eigenvectors. This is because the characteristic polynomial of A_n can have complex roots, which will result in complex eigenvalues. The corresponding eigenvectors will also be complex. However, A_n can also have real eigenvalues and eigenvectors, depending on the matrix's properties.

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