Prove:If $\lambda$ is an eigenvalue of $A$, then $\lambda^n$ is an eigenvalue of $A^n$.
I basically wanted to understand what's missing/incorrect in my proof. Please help me out. Many thanks.
Let $A$ be an nxn matrix with $x$ a nonzero vector and $\lambda$ a scalar s.t. $Ax = \lambda x$, where $\lambda$ is the eigenvalue and $x$ is the eigenvector w.r.t $\lambda$.
Now, post-multiplying both sides by $x^-1$, we have $$Ax.x^{-1} = λxx^{-1} $$ $$A = \lambda I\tag{1}$$, where $I$ is the identity matrix.
$(1)$ implies, to maintain the equality, if $A$ is pre-multiplied by $A$, then $\lambda$ needs to be premultiplied by $\lambda$. Thus, $$A^2 = \lambda^{2}I \tag{2}$$
It follows from $(2)$ that $$A^n = \lambda^nI$$
QED
Please comment on issues/problems with this proof. Will really appreciate. Thanks.
The flaw is in $x^{-1}.$ what is $x^{-1}$?
For example, for matrix $$A=\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}$$
what would your $x^{-1}$ be? $x$ is a vector in this case.
I am enclosing a proof that $\lambda^2$ is an eigenvalue for $A^2$.
Let $x$ be an eigenvector of $A$.
$$A^2x=A(Ax)=A(\lambda x)=\lambda Ax = \lambda (\lambda x) = \lambda^2 x$$
Hopefully you can see the proof for general $n$.