Relationship between eigenvalues of matrix squared Let $A$ be symmetric positive definite. Suppose $\lambda$ is the largest eigenvalue of $A^T A $. Then we know $\lambda(A^T A ) = \lambda(A^2 )$ 

Does it follow that $\lambda(A^2) = [\lambda(A)]^2 $ ?
Hint: If $A=U\Lambda U^T$ where $\Lambda$ is the diagonal matrix of eigenvalues and $U$ is the orthonormal matrix of eigenvectors, then $A^2=U\Lambda^2 U^T$ (can you prove this?). Let $\lambda_{max}(.)$ denote the largest eigenvalue. Thus $\lambda_{max}(A)=\lambda_{max}(\Lambda)$. Also, $\lambda_{max}(A^2)=\lambda_{max}(\Lambda^2)$. Now can you think of any situation where $\lambda_{max}(\Lambda^2)\leq\lambda_{max}(\Lambda)^2$ for any diagonal matrix $\Lambda$.
Hint: If $A=U\Lambda U^T$ where $\Lambda$ is the diagonal matrix of eigenvalues and $U$ is the orthonormal matrix of eigenvectors, then $A^2=U\Lambda^2 U^T$ (can you prove this?). Let $\lambda_{max}(.)$ denote the largest eigenvalue. Thus $\lambda_{max}(A)=\lambda_{max}(\Lambda)$. Also, $\lambda_{max}(A^2)=\lambda_{max}(\Lambda^2)$. Now can you think of any situation where $\lambda_{max}(\Lambda^2)\leq\lambda_{max}(\Lambda)^2$ for any diagonal matrix $\Lambda$.