Eigenvalues of power of matrices

Question

How come if $\lambda$ is an eigenvalue of $A$, then $\lambda^k$ is an eigenvalue of $A^k$?

And is its multiplicity necessarily the same?

Answer 1

Let $v$ be an eigenvector corresponding to $\lambda$. Then, by induction $$ A^k v = A^{k-1}(Av) = \lambda A^{k-1}v = \lambda^k v $$ hence $A^k$ has an eigenvalue $\lambda^k$. For the multiplicity note that $A = \binom{1\; 0}{0\; -1}$ has $\lambda = 1$ with multiplicity 1, but $A^2 = {\rm Id}$ has $\lambda^2 = 1$ with multiplicity 2.

Answer 2

$Av = \lambda v$.

$A^2v = \lambda Av$

$A^2v = \lambda (\lambda v) = \lambda ^2 v$

now continue by induction...

Answer 3

As $\lambda$ is an eigen value of $A$ so $\exists v\ne 0$ $Av=\lambda v$. Then $A^kv=A(\dots(A(A(A(A(v))\dots)\{\text{A is applied } $k$ \text{ times}\}=\lambda^kv$($\forall k\in N$ ) so $\lambda^k$ is an eigen value of $A^k$.

If $\lambda ^k\ne \omega^k$ for some eigen value $\omega $ (of $A$)then the multiplicity remains preserved.Otherwise it might not remain preserved.

Answer 4

Induction: suppose $\,Av=\lambda v\;,\;\;0\neq v\in V\;$ :

$$k=2\implies A^2v=A(Av)=A(\lambda v)=\lambda Av=\lambda(\lambda v)=\lambda^2v$$

Suppose for $\;k\;$, we shall prove for $\,k+1\,$ :

$$A^{k+1}=A(A^kv)\stackrel{\text{ind. Hyp.}}=A(\lambda^kv)=\lambda^kAv=\lambda^k(\lambda v)\ldots$$