Eigenvector of A_n

1. Dec 10, 2008

FourierX

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.

Last edited: Dec 10, 2008
2. Dec 10, 2008

gabbagabbahey

What is $A_n$?

3. Dec 10, 2008

FourierX

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. Dec 10, 2008

gabbagabbahey

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. Dec 10, 2008

FourierX

A_n is a n x n matrix with all elements 1.

A_nv = $$\lambda$$v

where,
$$\lambda$$ = eigvenvalue
???

6. Dec 10, 2008

gabbagabbahey

Yes. So does A_n times [1 1 1 ....1]T satisfy that condition?

7. Dec 10, 2008

FourierX

A_n * [1 1 1 1......1]T gives [n n n n.....n], doesn't it ?

8. Dec 10, 2008

gabbagabbahey

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?

9. Dec 10, 2008

FourierX

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. Dec 10, 2008

gabbagabbahey

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.

11. Dec 10, 2008

FourierX

Wow. Thanks a lot!

Could you also give me a slight hint on application of Eigenvalues and vectors?

12. Dec 10, 2008

HallsofIvy

Staff Emeritus
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.