If $A$ is symmetric with eigenvalue $\lambda$ then $\lambda^2$ is an eigenvalue of $A^TA$

Question

I am trying to show that if $A$ is symmetric and has an eigenvalue $\lambda$, then $\lambda^2$ is an eigenvalue of $A^TA$.

My thoughts:

Beginning with the definition, $A\vec{v}=\lambda\vec{v}. \tag{1}$ If $A$ is symmetric, then $A^T\vec{v}=\lambda\vec{v}. \tag{2}$ Multiplying $(2)$ with $(1)$, $$A^TA\vec{v}^2=\lambda^2\vec{v}^2\implies A^TA\vec{V}=\lambda^2\vec{V}, \ \ \text{where} \ \vec{V}=\vec{v}^2.$$ Hence $\lambda^2$ is an eigenvalue of $A^TA$ with corresponding eigenvector $\vec{V}$.

Answer 1

Let $v$ be an eigenvector of $A$ with corresponding eigenvalue $\lambda$. We then know that

\begin{equation} Av = \lambda v \; \; \; \textrm{(i)} \end{equation}

Since $A$ is symmetric, $A^T = A$, so \begin{equation} A^Tv = \lambda v \; \; \; \textrm{(ii)} \end{equation}

Hence,

\begin{equation} (A^TA)(v) = A^T(Av) = A^T(\lambda v) = \lambda (A^Tv) = \lambda^2 v \; \; \; \textrm{(iii)} \end{equation}

That is, $\lambda^2$ is an eigenvalue of $A^TA$, as desired.