diagonalising (I-X(X^TX)^-1X^T where X is rank n-k, a square matrix dimension nxn, into a matrix with n-k 1 along the diagonal, and then zeros.

Regarding the eigenvalues: note that $A = I_n - X(X^TX)^{-1}X$ is not only symmetric but also, as you've said, idempotent (i.e. $A^2 = A$). With that in mind: if $\lambda$ is an eigenvalue of $A$, then there must be an associated eigenvector $x$ (so that $x \neq 0$ and $Ax = \lambda x$), which means that $$ \lambda x = Ax = A^2x = A(Ax) = A(\lambda x) = \lambda Ax = \lambda^2 x. $$ So, we have $\lambda x = \lambda^2 x$. Because $x\neq 0$, it must be that $\lambda = \lambda^2$. This means that we must have $\lambda = 0$ or $\lambda = 1$.

The more general result is that if there is a polynomial $p(x)$ such that the matrix $A$ satisfies $p(A) = 0$, then it must hold that all eigenvalues of $A$ satisfy $p(\lambda) = 0$. In the case of an idempotent $A$, this holds with the polynomial $p(x) = x^2 - x$.

Regarding your manipulation of trace: it is correct to say that $$ \text{tr}(\underline{u}^T(I_n-X(X^TX)^{-1}X^T)\underline{u})=\text{tr}[(\underline{u}\underline{u}^T)(I_n-((X^TX)^{-1}X^TX))]. $$ From there, you seem to assume that $\operatorname{tr}(\underline{u}\underline{u}^T A) = \operatorname{tr}(\underline{u}\underline{u}^T) \operatorname{tr}(A)$ (where $A$ has its earlier definition), but this does not necessarily hold.