Thread: 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...

If v is an eigenvector of A with eigenvalue , then . Taking of both sides, so that and . In other words, if v is an eigenvector of A with eigenvalue then it is also an eigenvector of with eigenvalue . Enough?

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!

(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?

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, 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 A³ - I =

[7 48]
[0 ..7], which clearly has eigenvalue 7 = 2³ - 1.

Can't we just use the Cayley hamilton theorem and then the evalues of A are 1, -1, i, -i?

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?

[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".

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

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").