Let $A$ be an n×n matrix with eigenvalues $\lambda_i, i=1,2,\dots,n$. Then $\lambda_1^k,\dots,\lambda_n^k$ are eigenvalues of $A^k$.

  1. I was wondering if $\lambda_1^k,\dots,\lambda_n^k$ are all the eigenvalues of $A^k$?
  2. Are the algebraic and geometric multiplicities of $\lambda_i^k$ for $A^k$ same as those of $\lambda_i$ for $A$ respectively?


  • 4
    $\begingroup$ Note that eigenvalues can "merge", e.g., if $A$ has eigenvalues $1$ and $-1$, then $A^2$ has eigenvalue $1$. So the algebraic multiplicities can "combine". $\endgroup$ – Gerry Myerson Nov 21 '12 at 4:27
  • 1
    $\begingroup$ yes, they all should be eigenvalues of $A$. Simply look at the equation $Ax=\lambda x$. Since $A$ has $N$ of them, $A^{k}$ will also have the same $N$ eigen values (raised to the required power). Now only they can be the eigen values of $A^{k}$ since the characteristic equation is always a $N$-degree polynomial and can have only $N$ roots. (Just saw @GerryMyerson 's answer, see that one for algebraic multiplicity. $\endgroup$ – dineshdileep Nov 21 '12 at 4:27
  • $\begingroup$ Consider that $det(A^k) = det(A)^k$. If $f_1,\dots f_n$ are the e-values of $A^k$ then $det(A^k) = f_1f_2\cdots f_n$. On the other hand, we know $det(A) = \lambda_1\lambda_2 \cdots \lambda_n$ thus $\lambda_1^k\lambda_2^k \cdots \lambda_n^k = f_1f_2\cdots f_n$. Can we conclude for each $j$ there exists an $i$ such that $f_j = \lambda_i^k$? $\endgroup$ – James S. Cook Nov 21 '12 at 4:29
  • $\begingroup$ @GerryMyerson, dineshdileep, James S. Cook : Thanks! James: We cannot. $\endgroup$ – Tim Nov 21 '12 at 5:57

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.
  • $\begingroup$ Thanks! Does your reply apply to $k \in \mathbb{N}$ or $k \in \mathbb{Z}$? $\endgroup$ – Tim Nov 24 '12 at 18:03
  • $\begingroup$ "under the same height" means? $\endgroup$ – Tim Nov 24 '12 at 18:25
  • $\begingroup$ I originally wrote the solution for $k\in \mathbb{N}$ but it's easy to generalize to inverses when they exist. And what I mean by the "same height" is that if $i$ is the smallest integer for which $(A-\lambda I)^i$ annihilates $\mathbf{v}$ then $i$ is also the smallest integer for which $(A^k-\lambda I)^i$ annihilates $\mathbf{v}$. This means that a chain of generalized eigenvectors for $A$ remains a chain for $A^k$ and so $A^k$ shares a Jordan basis with $A$. $\endgroup$ – EuYu Nov 24 '12 at 19:38

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.

  • $\begingroup$ +1 Thanks! (1) What is the reference for the theorem? (2) Does the theorem or other similar one say something about algebraic and geometric multiplicities, and eigenvectors? $\endgroup$ – Tim Nov 24 '12 at 18:05
  • 3
    $\begingroup$ I should mention something here. Both this method and the one dineshdileep mentioned in the comments don't exactly show that the two spectra are the same. It only shows that the spectrum of $A$ is a subset of the spectrum of $A^k$. $\endgroup$ – EuYu Nov 24 '12 at 19:41
  • $\begingroup$ @Tim, I can't immediately find it on wikipedia or anything but the proof is really easy. Maybe that's why I can't find it. First establish that the sum, scalar multiplication, and powers of a matrix do the same to its spectrum. Then expand $f(x)$ into its Taylor series and apply the previous three lemmas and voila. Well-definition of f(A) and the convergence of the taylor expansion applied to A is also easily shown. $\endgroup$ – Fixed Point Nov 25 '12 at 10:11
  • $\begingroup$ @EuYu: the eigenvalues of $A^k$ are k-th power of the eigenvalues of $A$, so the eigenvalues of $A$ may not even be eigenvalues of $A^k$. Why "the spectrum of $A$ is a subset of the spectrum of $A^k$"? $\endgroup$ – Tim Nov 26 '12 at 3:49
  • $\begingroup$ @Tim I was being very careless with language there. I of course meant that if $\{\lambda_1,\ \cdots,\ \lambda_k\}$ was the spectrum of $A$, then $\{f(\lambda_1),\ \cdots,\ f(\lambda_k)\}$ is a subset of the spectrum of $f(A)$. $\endgroup$ – EuYu Nov 26 '12 at 4:20

If A is not diagonalizable, there exists nonsingular matrix P such that $\mathbf{P^{-1}AP}=\mathbf{J}=$ \begin{bmatrix} J_1 & 0& 0 & \dots & 0 \\ 0 & J2 & 0& \dots & 0 \\ \vdots & \vdots& \vdots & \vdots &\vdots\\ 0&0&0&\dots&J_l \end{bmatrix} with $J_i=\begin{bmatrix} \lambda_i & 1& 0 & \dots & 0 \\ 0 & \lambda_i & 1& \dots & 0 \\ \vdots & \vdots& \vdots & \vdots &\vdots\\ 0&0&\dots&0&\lambda_i \end{bmatrix}$. So $J_i^k=\begin{bmatrix} \lambda^k_i & C_k^1\lambda_i^{k-1}& \dots& \dots & C_k^{k-n_i-1}\lambda_i^{k-n_i-1} \\ 0 & \lambda^k_i & \dots& \dots & \vdots \\ \vdots & \vdots& \vdots & \vdots &\vdots\\ 0 & 0 & \dots& \dots & C_k^1\lambda_i^{k-1}\\ 0&0&\dots&\dots&\lambda^k_i \end{bmatrix}$

Then $\mathbf{J}^k=\mathbf{P}^{-1}\mathbf{A}^k\mathbf{P}$, so $\mathbf{J}^k=\begin{bmatrix} J^k_1 & 0& 0 & \dots & 0 \\ 0 & J^k_2 & 0& \dots & 0 \\ \vdots & \vdots& \vdots & \vdots &\vdots\\ 0&0&0&\dots&J^k_l \end{bmatrix}$ and $\mathbf{A}^k$ have the same eigenvalues. Since the size of each Jordan block does not change, the algebraic multiplicities do not change.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.