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Eigenvector of A_n

 
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Dec10-08, 12:29 AM   #1
 

Eigenvector of A_n


1. The problem statement, all variables and given/known data

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



2. Relevant equations

N/A

3. 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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Dec10-08, 12:32 AM   #2
 
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What is [itex]A_n[/itex]?
 
Dec10-08, 12:38 AM   #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]
 
Dec10-08, 12:42 AM   #4
 
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Eigenvector of A_n


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?
 
Dec10-08, 12:47 AM   #5
 
A_n is a n x n matrix with all elements 1.



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

where,
[tex]\lambda[/tex] = eigvenvalue
???
 
Dec10-08, 12:50 AM   #6
 
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Yes. So does A_n times [1 1 1 ....1]T satisfy that condition?
 
Dec10-08, 12:58 AM   #7
 
A_n * [1 1 1 1......1]T gives [n n n n.....n], doesn't it ?
 
Dec10-08, 01:00 AM   #8
 
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Actually it gives [n n n n.....n]T

Is that a scalar multiple of [1 1 1 1......1]T? If so, what is the eigenvalue?
 
Dec10-08, 01:04 AM   #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 ?
 
Dec10-08, 01:07 AM   #10
 
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Yup

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.
 
Dec10-08, 01:09 AM   #11
 
Wow. Thanks a lot!

Could you also give me a slight hint on application of Eigenvalues and vectors?
 
Dec10-08, 06:56 PM   #12
 
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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.
 
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