How will matrix $A^n$ affect the original eigenvector and eigenvalue?

Question

For example, a matrix $A$ has three distinct eigenvalues and has 3 eigenvectors $v_1,v_2,v_3$ corresponding to the three distinct eigenvalues.

So, am I right to say $A^4$ will result in eigenvalue^4 and the eigenvectors remain unchanged?

Answer 1

The claim that the eigenvectors remain unchanged can be interpreted in two different ways, one of which is true, while the other is false.

The first interpretation is

If $v$ is an eigenvector of $A$, then $v$ is an eigenvector of $A^4$.

This claim is true, and easily verified:

If $v$ is an eigenvector of $A$, then there exists a value $\lambda$ such that $Av=\lambda v$. But then $A^4v = A^3Av = A^3(\lambda v) = \lambda A^3v = A^2 A v = \ldots = \lambda^4 v$. Therefore $v$ is also an eigenvector of $A^4$, with the corresponding eigenvalue $\lambda^4$.

The second interpretation is:

$v$ is an eigenvalue of $A^4$ iff $v$ is an eigenvalue of $A$.

That claim is false. A simple counterexample is $A=\pmatrix{1&0\\0&-1}$. Up to a scalar factor, the only eigenvectors of this matrix are $\pmatrix{1\\0}$ and $\pmatrix{0\\1}$. But $A^4$ is the identity matrix, and all non-zero vectors are eigenvectors of the identity.

With the eigenvalues it is even more complicated, as here the claim may depend on whether you consider real or complex vector spaces/matrices.

For example, consider the matrix $A=\pmatrix{2&0&0\\0&0&-1\\0&1&0}$. In the real numbers, the only eigenvalue is $2$. But $A^4$ has eigenvalues $2^4=16$ and $1$.

On the other hand, in the complex numbers, this matrix has the additional eigenvalues $i$ and $-i$, whose fourth powers are, indeed, both $1$.