# Linear Algebra: Eigenvalues question

• Jan 17th 2009, 11:34 PM
openustudent
Linear Algebra: Eigenvalues question
Given a matrix A over field F of order 5x5, with tr(A)=0, and p(A)=1, find the eigenvalues of A.

If the question would have been over C/R I would have found an answer, but this F field thing is too much for me...

Thanks!
• Jan 18th 2009, 06:57 AM
Isomorphism
Quote:

Originally Posted by openustudent
Given a matrix A over field F of order 5x5, with tr(A)=0, and p(A)=1, find the eigenvalues of A.

If the question would have been over C/R I would have found an answer, but this F field thing is too much for me...

Thanks!

It will help if you tell us what your notation means. what does p(A) = 1 mean?
• Jan 18th 2009, 08:47 AM
openustudent
p(A) is the rank of the matrix A (in our case - 1) and tr(A) is the trace of the matrix A.
• Jan 18th 2009, 09:09 AM
Isomorphism
Quote:

Originally Posted by openustudent
Given a matrix A over field F of order 5x5, with tr(A)=0, and p(A)=1, find the eigenvalues of A.

If the question would have been over C/R I would have found an answer, but this F field thing is too much for me...

Thanks!

Since the rank is 1, there exists vectors $\displaystyle x$ and $\displaystyle y$ such that $\displaystyle A = xy^T$(Prove this!)

This matrix has atmost one non zero eigenvalue. Clearly $\displaystyle x$ is an eigen vector corresponding to the non-zero(if it is...) eigenvalue $\displaystyle x^T y$.

But observe that $\displaystyle \text{Tr} (xy^T) = x^Ty$, thus all eigenvalues of A are 0.
• Jan 18th 2009, 09:20 AM
openustudent
There are two statements though that I did not understand:

Quote:

This matrix has atmost one non zero eigenvalue
How come ?

Quote:

Clearly http://www.mathhelpforum.com/math-he...155c67a6-1.gif is an eigen vector corresponding to the non-zero(if it is...) eigenvalue http://www.mathhelpforum.com/math-he...29808f9b-1.gif.
Why is this?
• Jan 18th 2009, 09:31 AM
Isomorphism
Quote:

Originally Posted by openustudent
How come ?
Why is this?

Observe that $\displaystyle Ax = (xy^T)x = x(y^Tx) = (y^Tx)x$. If you still cant see it, observe that $\displaystyle y^Tx$ is a scalar.
Let us say it had an eigenvector z, corresponding to a eigen value $\displaystyle \alpha$ ,then $\displaystyle Az = \alpha z \implies (xy^T)z = \alpha z \implies (y^T z) x = \alpha z$ , Thus if $\displaystyle \alpha \neq 0$, z is in the span of x.