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 A^{3}-A^{-1}?

I know that the eigenvalues of A^{3} 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 A^{3} 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 A^{3}?

In this case I would have to do a lot of matrix multiplication which takes a lot of time...

Re: Eigenvalues of A^3-A^(-1)

If v is an eigenvector of A with eigenvalue $\displaystyle \lambda$, then $\displaystyle Av= \lambda v$. Taking $\displaystyle A^{-1}$ of both sides, $\displaystyle A^{-1}(Av)= A^{-1}\lambda v$ so that $\displaystyle v= \lambda A^{-1}v$ and $\displaystyle A^{-1}v= (1/\lambda)v$. In other words, if v is an eigenvector of A with eigenvalue $\displaystyle \lambda$ then it is also an eigenvector of $\displaystyle A^{-1}$ with eigenvalue $\displaystyle 1/\lambda$. Enough?

Re: Eigenvalues of A^3-A^(-1)

So if I have eigenvalues following eigenvalues for A: 3 and 7.

For A^{3} I have eigenvalues of 9 and 27, than for A^{3}-A^{-1} I have eigenvalues:

9-1/3 and 27-1/7

right?

Thank you!

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 A^{3}-I

Can I simply take the eigenvalue of A^{3} and subtract 1?

Re: Eigenvalues of A^3-A^(-1)

well, let's see if that works:

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

(A^{3} - I)(v) = A^{3}(v) - I(v) = A^{2}(Av) - v

= A^{2}(λv) - v = A(A(λv)) - v = A(λ(A(v))) - v = A(λ(λv)) - v = A(λ^{2}v) - v = λ^{2}(A(v)) - v

= λ^{2}(λv) - v = λ^{3}v - v = (λ^{3} - 1)v.

in other words, for all integers k, A^{k} has eigenvalue λ^{k} (even when k = 0) (if A^{k} 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 A^{3} - I =

[7 48]

[0 ..7], which clearly has eigenvalue 7 = 2^{3} - 1.

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?

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?

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".

Re: Eigenvalues of A^3-A^(-1)

can you please tell me how 48 comes in the matrix without using matrix multiplication.

Re: Eigenvalues of A^3-A^(-1)

A^{3} *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").