Prove that the trace of a normal matrix is equal to the sum of the eigenvalues Prove that the trace (main diagonal sum) of a normal matrix is equal to the sum of the eigenvalues.
Note: For a matrix A to be normal we must have AA*=A* A where A* is the Hermitian Conjugate
I am working on this proof and not sure how to proceed other than writing the trace out for a normal matrix but a hint or some guidance would be greatly appreciated.
 A: Hints:
If $A$ is normal, it is orthogonally diagonalizable, that is, there exists unitary $Q$ ($Q^*Q = QQ^* = I$) so that 
$$
D = Q^*AQ
$$
is a diagonal matrix, with the eigenvalues of $A$ along the diagonal.
Then, trace properties (like $\textrm{tr}(A^TB) = \textrm{tr}(AB^T)$ will finish the job.
A: Bring $A$ to Jordan form: $A=SJS^{-1}$. Since tr$(AB)=$tr$(BA)$, you find:
tr$A=$ tr$(SJS^{-1})=$ tr $(S^{-1}SJ)=$ tr $J=\sum_{i=1}^n \lambda_i$
A: Let $A$ be a normal matrix, then there is a basis for $V$ of eigenvectors of $A$. Let $D = P^{-1} A P$ a diagonal matrix, where $P$ is the matrix that has the eigenvectors of $A$ on the columns. So $D = diag(\lambda_1,\ldots,\lambda_n)$, where $\lambda_i$ is the eigenvalue associated with the $i$-th vector of the basis.
Then $tr(D) = \sum \lambda_i$. But trace is invariant by similarities and then $tr(A) = \sum \lambda_i$
Other way of seeing it:
The characteristic polynomial for $A$, $p_T(A) = \det(xI - A)$ has all eigenvalues as roots, and as $A$ is diagonalizable, the sum of all eigenvalues is minus the coefficient of the term that multiplies $x^{n-1}$. But as $\det(xI-A) = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) (xI-A)_{1\sigma(1)}\ldots (xI-A)_{n\sigma(n)}$, we need to choose $\sigma(i)=i$ for at least $n-1$ values of $\sigma$, but the only permutation that does that is $\sigma=Id$, hence the only possible term that has $x^{n-1}$ is $(xI-A)_{11}\ldots(xI-A)_{nn}$ = $(x - a_{11})\ldots (x-a_{nn})$, but the coeficient of $x^{n-1}$ in the expansion of $(x - a_{11})\ldots (x-a_{nn})$ is precisely $ - (a_11 + \ldots + a_nn) = -tr(A)$
