# Finding the eigenvalues and eigenvectors of $A^{n}$

Find the eigenvalues and eigenvectors of $A^{5}$ for $A = \begin{bmatrix} 0&0&-1 \\-1&1&-1 \\ 1&-1&0\end{bmatrix}$. How many eigenspaces does it have? What is the dimension of each eigenspace and what are the bases vectors for it?

I multiplied $A$ by $A$ and found $A^{2}$, then multiplied $A$ by $A^{2}$ and found $A^{3}$, and multiplied $A^{2}$ by $A^{3}$ and found $A^{5}$. From the characteristic equation I found eigenvalues 0, 32 and -1. And I found eigenvectors and then the bases for eigenspaces from the equation (λI - A)x = 0. Let me get down to business. Eigenvalues' being fifth powers of integers made me get the feeling that it wasn't the shortest way to solve the problem. If it wasn't, what is the thing that I don't know?

• If A has eigenvalue x then A^5 has eigenvalue x^5. – SebiSebi Dec 23 '14 at 21:42
• have you found all the eigenvalues? look for the characteristic equation of $A.$ you may also note that sum of the first and second column is zero so $0$ is an eigenvalue corresponding eigenvector is $(1,1,0)^T$ same will be true for $A^5$ – abel Dec 23 '14 at 22:40

If $\lambda$ is an eigenvalue of $A$, then $\lambda^{5}$ is an eigenvalue of $A^{5}$. The eigenvectors are the same.
So we have $Av = \lambda v$. And $A^{2}v = AAv = A(\lambda v) = \lambda Av = \lambda^{2}v$. By induction, $\lambda^{n}$ is an eigenvalue of $A^{n}$ and $v$ is an eigenvector of each positive power of $A$.