What are the Eigenvalues of $A^2?$ If $a$ is an eigenvalue of $A$ then $A^2$ has an eigenvalue $a^2.$ But what about the multiplicities? I mean is it possible to have $1,1,2$ as the eigenvalues of a $3\times 3$ matrix $A$ and $1,4,8$ as the eigenvalues of $A^2?$
 A: Let 
$$P(\lambda)=\det(\lambda I-A)=(\lambda-\lambda_1)\cdots(\lambda-\lambda_n)$$ 
be the characteristic polynomial for $A$ Then 
$$\begin{align}
\det(\lambda I+A)&=(-1)^n\det(-\lambda I-A)\\
&=(-1)^nP(-\lambda)\\
&=(-1)^n(-\lambda-\lambda_1)\cdots(-\lambda-\lambda_n)\\
&=(\lambda+\lambda_1)\cdots(\lambda+\lambda_n)
\end{align}$$ 
Now if $Q(\lambda)=\det(\lambda I-A^2)$ is the characteristic polynomial for $A^2$, then
$$\begin{align}
Q(\lambda^2)
&=\det(\lambda^2I-A^2)\\
&=\det(\lambda I-A)\det(\lambda I+A)\\
&=(\lambda-\lambda_1)\cdots(\lambda-\lambda_n)(\lambda+\lambda_1)\cdots(\lambda+\lambda_n)\\
&=(\lambda^2-\lambda_1^2)\cdots(\lambda^2-\lambda_n^2)
\end{align}$$
which gives us an identity for the polynomial $Q$. It follows that $Q(\lambda)=(\lambda-\lambda_1^2)\cdots(\lambda-\lambda_n^2)$. Thus the eigenvalues of $A^2$ cannot be anything other than the squares of the eigenvalues of $A$.
A: If $A$ is a square matrix with eigenvalues $\lambda$, we can apply induction on the size of the matrix to show that the eigenvalues of $A^{k}$ are $\lambda^{k}$ for any positive integer $k$. Hence, the eigenvalues of $A^{2}$ are exactly $\lambda^{2}$ (the squares of the eigenvalues of $A$). See here: Show that $A^k$ has eigenvalues $\lambda^k$ and eigenvectors $v$.
A: If $a$ is an eigenvalue of $A$ with multiplicity $k$, then $a^2$ is an eigenvalue of $A^2$ with multiplicity at least $k$. So, the answer is negative: if $1$ is an eigenvalue of $A^2$ and its multiplicity as at least $2$, then $4$ and $8$ can't be eigenvalues too.
