If $\lambda$ is an eigenvalue of $A^2$, then either $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ is an eigenvalue of $A$ $A$ is an $n\times n$ matrix of complex numbers. Prove that if $\lambda$ is an eigenvalue of $A^2,$ then $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ is an eigenvalue of $A.$

If $\lambda$ is an eigenvalue of $A^2,$ we have $\lambda\alpha=A^2\alpha$ for some $\alpha.$ Then how can we find a $\beta$ s.t. $\sqrt{\lambda}\beta=A\beta?$
 A: First note that: $$A^2 - \lambda I = (A-\sqrt\lambda I)(A+\sqrt\lambda I)$$ Let $v$ be an eigenvector of $A^2$ with eigenvalue $\lambda$. We can use $v$ to find an explicit eigenvector of $A$ with eigenvalue that is either $\sqrt\lambda$ or $-\sqrt\lambda$. 
Since $(A^2-\lambda I)v = 0$, we must have either $(A+\sqrt\lambda I)v = 0$, in which case $v$ is also an eigenvector of $A$ with eigenvalue $-\sqrt\lambda$, or $(A+\sqrt\lambda I)v \neq 0$, in which case we set $ u:=(A+\sqrt\lambda I)v$, and note that $u$ is an eigenvector of $A$ with eigenvalue $\sqrt\lambda$, since $(A-\sqrt\lambda I)u=0$.
A: Since $\lambda$ is an eigenvalue of $A^2$, we know that
$$\det (A^2 - \lambda I) = 0$$
From here we conclude that
$$\det (A^2 - \lambda I) = \det((A - \sqrt{\lambda}I)(A + \sqrt{\lambda}I)) = \det(A - \sqrt{\lambda}I) \times\det ( A + \sqrt{\lambda}I)= 0$$
Hence $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ is an eigenvalue of $A$.
A: It follows from the Jordan decomposition that if $v_{\lambda^2}$ is an eigenvector of $A^2$ with eigenvalue $\lambda^2$, then it is either an eigenvector of $A$ corresponding to the eigenvalue $\lambda$ resp. $-\lambda$, or a combination $v_{\lambda}+v_{-\lambda}$. To see this, just square the Jordan block
$$
\begin{pmatrix}
\lambda & 1 & 0 & \ldots & 0\\
0 & \lambda & 1  & \ldots & 0\\
& \ldots & & & & \\
0 & \ldots & & 0 & \lambda
\end{pmatrix}
$$
to get a matrix with eigenvalue $\lambda^2$ and only one eigenvector $(1,0,\ldots, 0)$. 
A: Hint: use Jordan canonical form of $A$. 
