Eigenvalues and power of a matrix

Solution 1:

The powers of the eigenvalues are indeed all the eigenvalues of $A^k$. I will limit myself to $\mathbb{R}$ and $\mathbb{C}$ for brevity.

The algebraic multiplicity of the matrix will indeed be preserved (up to merging as noted in the comments). An easy way to see this is to take the matrix over $\mathbb{C}$ where the minimal polynomial splits. This means that the matrix is triangularizable. The spectrum of the matrix appear on the diagonals of the triangularized matrix and successive powers will alter the eigenvalues accordingly.

Geometric multiplicity is not as easy to answer. Consider a nilpotent matrix $N$ which is in general not diagonalizable over $\mathbb{R}$ or $\mathbb{C}$. But eventually, we will hit $N^k = 0$ which is trivially diagonal.

This can be generalized: The geometric multiplicity of $0$ will always increase to the algebraic multiplicity of $0$ given enough iterations, this is because the zero Jordan blocks of a matrix is nilpotent and will eventually merge with enough powers.

For non-zero eigenvalues, this is a bit different. Consider the iteration of generalized eigenspaces for $A^k$ given by $$(A^k-\lambda^kI)^i\mathbf{v}$$ for successive powers of $i$. The difference of powers above splits into the $k$th roots of $\lambda^k$ given as $$[P(A)]^i(A-\lambda I)^i\mathbf{v}$$ where we collect the other $k$th roots into a polynomial $P(A)$. Notice that if the matrix does not contain two distinct $k$th roots of $\lambda^k$ then $P(A)$ is in fact invertible. This means that $$[P(A)]^i(A-\lambda I)^i\mathbf{v} = \mathbb{0} \iff (A-\lambda I)^i\mathbf{v} = \mathbb{0}$$ Note that this remains true for each generalized eigenspace even if there are distinct eigenvalues which share a $k$th power because the mapping $(A-\lambda I)$ restricted to the generalized eigenspace of another eigenvalue is injective.

This means that generalized eigenvectors of $A$ remain generalized eigenvectors of $A^k$ under the same height. More precisely, this implies that the structures of the generalized eigenspaces are identical and so the matrix power $A^k$ has the same Jordan form for that eigenvalue as $A$. It follows that the dimension of the regular eigenspace is also retained, so the geometric multiplicity is stable.

In Summary:

1. The spectrum is changed as expected to the $k$th powers.
2. The algebraic multiplicities of the eigenvalues powers are retained. If eigenvalues merge, then their multiplicities are added.
3. The geometric multiplicity of $0$ will increase to the algebraic multiplicity. The geometric multiplicities of other eigenvalues are retained. Again if there is merging, then the multiplicities are merged.

Solution 2:

I feel like I should mention this theorem. Forgot its name but I think its one of the spectral theorems. It says that if $\lambda$ is an eigenvalue for a matrix A and $f(x)$ is any analytic function, then $f(\lambda)$ is an eigenvalue for $f(A)$. So even $\sin(A)$ will have $\sin(\lambda)$ as its eigenvalues.

In your case, just take $f(x)=x^k$ and then apply it to all of the eigenvalues. So yes, $\lambda_n^k$ are all of the eigenvalues.