Suppose there is a square binary matrix (Adjacency matrix of a graph), $A$.
I got that, the matrices, $A^2$ and $A^3$ are distinct but the set of eigenvalues are same for $A^2$ and $A^3$. It is to be noted that the set of eigenvalues of $A$ is different from the same of $A^2$ and $A^3$. Other powers of $A$ are same as $A^3$.
What does the above result interpret?
Please let me know.
Thanks in advance!
On
Actually, if $\lambda$ is an eigenvalue of $A$, then $\lambda^{n}$ is an eigenvalue of $A^{n}$. The eigenvectors are invariant for powers of a matrix.
We note that if $\lambda$ is an eigenvalue of $A$, then $A^{2}v = A(\lambda * v) = \lambda * (Av) = \lambda * (\lambda * v) = \lambda^{2} v$. We simply apply the identity that $Av = \lambda * v$. Of course, I am assuming $v$ to be the corresponding eigenvector.
For any square matrix $A$, the eigenvalues of $A^k$ (counted by algebraic multiplicity) are the $k$'th powers of the eigenvalues of $A$. I'm having trouble thinking of a case where the eigenvalues of $A^3$ and $A^2$ are the same but are not the same as the eigenvalues of $A$. Certainly all eigenvalues would be either $0$ or have absolute valu e $1$. Moreover, if $A$ has rational entries and you have an irrational eigenvalue, it will be algebraic and its conjugates will also be eigenvalues.