# Eigenvalues of A^3-A^(-1)

• Jun 7th 2012, 01:47 PM
infernalmich
Eigenvalues of A^3-A^(-1)
I have to find the Eigenvalues of a Matrix A. This is pretty easy but the second part of the question is:
What are the corresponding eigenvalues and eigenvectors for A3-A-1?
I know that the eigenvalues of A3 are equal to the third power of the eigenvalues of A, so this is no problem.

If the Eigenvalues of A are -2 and 3 like in my case, the Eigenvalues for A3 are -8 and 27.

But I don't know how to deal with the subtraction of the inverse, is there any rule or do I have to take the inverse of the matrix manually and subtract it from A3?
In this case I would have to do a lot of matrix multiplication which takes a lot of time...
• Jun 7th 2012, 02:21 PM
HallsofIvy
Re: Eigenvalues of A^3-A^(-1)
If v is an eigenvector of A with eigenvalue $\lambda$, then $Av= \lambda v$. Taking $A^{-1}$ of both sides, $A^{-1}(Av)= A^{-1}\lambda v$ so that $v= \lambda A^{-1}v$ and $A^{-1}v= (1/\lambda)v$. In other words, if v is an eigenvector of A with eigenvalue $\lambda$ then it is also an eigenvector of $A^{-1}$ with eigenvalue $1/\lambda$. Enough?
• Jun 8th 2012, 07:55 AM
infernalmich
Re: Eigenvalues of A^3-A^(-1)
So if I have eigenvalues following eigenvalues for A: 3 and 7.

For A3 I have eigenvalues of 9 and 27, than for A3-A-1 I have eigenvalues:

9-1/3 and 27-1/7

right?

Thank you!
• Jun 8th 2012, 01:47 PM
infernalmich
Re: Eigenvalues of A^3-A^(-1)
(s there a rule for the Inverse too?

For example if I have to find the eigenvalue of A3-I
Can I simply take the eigenvalue of A3 and subtract 1?
• Jun 8th 2012, 02:59 PM
Deveno
Re: Eigenvalues of A^3-A^(-1)
well, let's see if that works:

suppose an eigenvalue of A is λ, with eigenvector v.

(A3 - I)(v) = A3(v) - I(v) = A2(Av) - v

= A2(λv) - v = A(A(λv)) - v = A(λ(A(v))) - v = A(λ(λv)) - v = A(λ2v) - v = λ2(A(v)) - v

= λ2(λv) - v = λ3v - v = (λ3 - 1)v.

in other words, for all integers k, Ak has eigenvalue λk (even when k = 0) (if Ak is even defined for k < 0, that is, if A is invertible).

for example, let A be the matrix:

[2 4]
[0 2], which has eigenvalue 2. then A3 - I =

[7 48]
[0 ..7], which clearly has eigenvalue 7 = 23 - 1.
• Jun 8th 2012, 03:54 PM
Electric
Re: Eigenvalues of A^3-A^(-1)
Can't we just use the Cayley hamilton theorem and then the evalues of A are 1, -1, i, -i?
• Jun 8th 2012, 11:05 PM
infernalmich
Re: Eigenvalues of A^3-A^(-1)
Thank you for your explaination, but:
you got the following Matrix for A^3-I
[7 48]
[0 ..7]

if I calculate it I get:
[7 64]
[0 ..7], why do you have the 48? Is there any rule I didn't took into consideration?
• Jun 9th 2012, 02:40 AM
Deveno
Re: Eigenvalues of A^3-A^(-1)
[2 4]3
[0 2] =

[2 4][2 4][2 4]
[0 2][0 2][0 2] =

[4 16][2 4]
[0 ..2][0 2] = (in the upper-right we have 2*4 + 4*2 = 8 + 8 = 16)

[8 48]
[0 ..8] (in the upper-right we have 4*4 + 16*2 = 16 + 32 = 48)

you don't just "cube each entry".
• Jun 9th 2012, 06:11 AM
saravananbs
Re: Eigenvalues of A^3-A^(-1)
can you please tell me how 48 comes in the matrix without using matrix multiplication.
• Jun 9th 2012, 07:01 AM
Deveno
Re: Eigenvalues of A^3-A^(-1)
A3 *is* matrix multiplication (it's A times A times A). there's no simpler way to describe the entries of powers of a matrix, except by doing the multiplication (unless the matrix is diagonal, or of some other "special form").