I'm trying to prove that a unitary matrix can be diagonalized using an orthonormal basis of eigenvectors, and that the eigenvalues are on the unit circle. So far, I have been able to show that if A is unitary, then A is unitarily similar to T, which is unitary and upper-triangular. Basically, I need to show that T is diagonal, but I'm not sure how to do this.
Additionally, I want to prove that a matrix is positive definite iff it can be diagonalized using an orthonormal basis of eigenvectors, and that the eigenvalues are real and positive. I've already written a similar proof for Hermitian matrices, with the eigenvalues real (but not necessarily positive). However, I'm having trouble seeing where to start with this one.
For your first question, partition $T$ as $\pmatrix{a& b^\ast\\ 0&C}$, where $a$ is a scalar. Then $I=TT^\ast=\pmatrix{|a|^2+\|b\|^2& b^\ast C^\ast\\ Cb&CC^\ast}$. It follows that $C$ is unitary, $b=0$ and $|a|=1$. Yet $C$ is upper triangular. Therefore, by mathematical induction, $C$ is diagonal. Hence $T$ is diagonal.
For your second question, note that every positive definite matrix is, by definition, Hermitian.