Calculus eigenvalues

Question

Supppose A is an invertible n×n matrix and v is an eigenvector of A with associated eigenvalue 4. Convince yourself that v is an eigenvector of the following matrices, and find the associated eigenvalues:

A⁸ =

A^-1 =

A+4I_n =

-3A =

Answer 1

So this was actually a neat fact that I discovered in messing around with eigenvectors and eigenvalues

When you raise the original matrix A by some power, say we call it 'a', then the eigenvector remains unchanged by the associated eigenvalue with change like so:

λ' ( the new eigenvalue for A^a ) = λ^a (the λ here is the λ eigenvalue from the original, unaltered matrix A)

So for, A⁸ and A^-1, its respective eigenvalues will be:

λ₁ = 4⁸ = 65536 and λ₂ = 4^-1 = 1/4 = 0.25

Now when it comes to adding (or substracting) a multiple of an identity matrix, the following will be how you'd make the adjustment to the new eigenvalue:

λ' = λ + b

Where 'b' is from: A + b * I_n

This is due to the fact that the change from that identity matrix can be "absorbed" into a pseudo-eigenvalue so that the solution process is the exact same like the prior eigenvalues, and then shift the pseudo-eigenvalues by that amount b

So for b = +4,

λ₃ = 4 + 4 = 8

And lastly, when you scale the original matrix by some constant, say we call it 'h', then the eigenvalues will scale by that same 'h' factor. It's similar to the pseudo-eigenvalue approach as above but involves setting the pseudo-eigenvalue to: λ' = λ / h

(this is because in the final step, we want to multiply the eigenvalues that we got before by h, assuming h was a whole number; 'h' can techincally be whatever value the problem states but I am working off the starting fact that 'h' was a whole number to arrive at my pseudo formula stated above)

λ₄ = -3 * 4 = -12

I hope this helps! Message me in the comments if you have any questions, comments, or concerns!