Show that the $\text{Tr}(A)^2 = \text{Tr}(A^2)+\text{Sum of Eigenvalues} $ Let $A$ be a square $ m \times m $ with eigenvalues $\lambda_{i},...,\lambda_{m}$. Show that:
$$
[\text{Tr}(A)]^{2} =\text{Tr}(A^{2}) + \sum_{i \neq j} \lambda_{i}\lambda_{j}
$$
Here is my attempt:
LHS
$$
[\text{Tr}(A)]^{2} = \sum_{i =1}^{m} \lambda_{i}\sum_{j =1}^{m}\lambda_{j} = \sum_{i =1}^{m}\sum_{j =1}^{m} \lambda_{i}\lambda_{j}
$$
RHS
It can be shown that $\lambda_{i}^{2}$ is an eigenvalue of $A^{2}$ so:
$$
\text{Tr}(A^{2}) + \sum_{i \neq j} \lambda_{i}\lambda_{j} = \sum_{i =1}^{m} \lambda_{i}^{2} + \sum_{i \neq j} \lambda_{i}\lambda_{j}
$$
$$
= \sum_{i =1}^{m}\sum_{j =1}^{m} \lambda_{i}\lambda_{j}
$$
Does this make sense?
 A: This is perfectly correct. $$\\$$
As a plus, you can show the "it can be shown that", in a one-line proof:
If $\lambda$ is an eigenvalue of $A$ with eigenvector $v$,
$$ A^2 v = A(Av) = A(\lambda v) = \lambda (Av)= \lambda^2 v $$
A: If $\lambda_1,\dots,\lambda_m$ are distinct, this is totally fine. If not, you just need to make sure you're counting them the right number of times, i.e., that $\lambda_1,\dots,\lambda_m$ are the roots of the characteristic polynomial with multiplicity. And then you should understand that these multiplicities work right when you square a matrix. For example, suppose $A$ is $3\times3$ and has eigenvalues $1$ and $-1$ with multiplicity $2$, i.e., "eigenvalues" $1$, $-1$, and $-1$. As you say, we can conclude that $1$ is an eigenvalue of $A^2$, but we need to be able to conclude that $1$ is an eigenvalue with multiplicity $3$. A priori, $A^2$ could have eigenvalues $1$, $\pi$, and $e$. 
Also, if these eigenvalues do not lie in the underlying field, you just need to make sure you're still OK calling them eigenvalues even if they don't have eigenvectors. For example, if we are working with real matrices (matrices over $\mathbb{R}$), we need to make sure we're good with matrices like 
$$\begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix}$$
which have no eigenvectors over $\mathbb{R}$. This is another reason to instead talk about roots of the characteristic polynomial.
