# eigenvalue/eigenvector

• Nov 20th 2007, 06:51 AM
zbrn
eigenvalue/eigenvector
Show that if a matrix A has an eigenvector x corresponding to the eigenvalue h, then x is also an eigenvector of the matix A^2 and state the corresponding eigenvalue.
Write an analogous result for A^n.

I'm afraid this kind of has me stumped. I have managed to do a couple of questions on eigenvalues and eigenvectors, but I can't apply that knowledge to this problem. :(

Anyone help?!? :o
• Nov 20th 2007, 08:36 AM
ThePerfectHacker
If $k$ is an eigenvalue of $A$ then it means $A\bold{x} = k\bold{x}$ has a non-trivial solution $\bold{x}$.
But that means $A^2\bold{x}=A(A\bold{x})=A(k\bold{x}) = k(A\bold{x}) = k(k\bold{x})=k^2\bold{x}$.
Thus, $\bold{x}$ is an eigenvector of $A^2$ corresponding with eigenvalue $k^2$.
• Nov 20th 2007, 01:11 PM
zbrn
Thank you for the help ThePerfectHacker !

I am wondering if someone can help me with another similar problem. Rather than start a new thread I'll ask here. Please have mercy, I am not sure how to format the math correctly. :D Here goes...

I have a matrix:
1 -1
1 0

I have got the eigenvalues to be: (1 +/- sqrt3) / 2
or e^(i.pi/3) & e^-(i.pi/3)

I must now show what the corresponding eigenvectors are, and I am not sure of the answer.
• Nov 20th 2007, 02:41 PM
kalagota
Quote:

Originally Posted by zbrn
Thank you for the help ThePerfectHacker !

I am wondering if someone can help me with another similar problem. Rather than start a new thread I'll ask here. Please have mercy, I am not sure how to format the math correctly. :D Here goes...

I have a matrix:
1 -1
1 0

I have got the eigenvalues to be: (1 +/- sqrt3) / 2
or e^(i.pi/3) & e^-(i.pi/3)

I must now show what the corresponding eigenvectors are, and I am not sure of the answer.

let k be an eigenvalue of the matrix A..
then the corresponding eigenvector are the matrices of the form

$(kI - A) X = 0$ and solve for the x's..
note, they may not be unique. in fact, they form a set called the Eigenspace with respect to k.. (note again that as the name implies, the set is a space..)