If M is hermitian then M* = M, and M is normal. A skew hermitian matrix is also normal.
Let M be an orthonormal matrix, hence M and M* are inverses. A left inverse is a right inverse, so M and M* commute, and M is a normal matrix.
If M is orthogonal, them M times its tranjugate, or the other way around, produces a diagonal matrix whose entries are the lengths of the vectors of M. Once again M is normal.
A diagonal matrix is orthogonal, so it is normal. Or you can just multiply D×D* and D*×D and verify it yourself.
Let M be diagonalizable, as described in an earlier section. In other words, PM/P is a diagonal matrix, where P is nonsingular. Also, assume P is unitary, so that M is unitarily similar to a diagonal matrix. Remember that Q, the inverse of P, is also the tranjugate of P.
Consider M times its tranjugate.
MM* =
qdp × qd*P =
qdd*P =
qd*dp =
qd*P × qdp =
M*M
This shows M is normal. Conversely, if M is normal and P is unitary, then PMQ is also normal. So apply Schur's theorem, making PMQ an upper triangular matrix T. Thus T commutes with its tranjugate. We want to show that T is in fact diagonal.
Let L be the tranjugate of T, so that L is lower triangular. Let c be the upper left entry of L; hence the conjugate of c is the upper left entry of T. Consider L*T. The top row of the product is c times the top row of T. In fact the upper left entry is c times its conjugate.
Now look at the upper left entry of T*L. It has to be the same, namely c times its conjugate. This is satisfied by T1,1×L1,1. Therefore the rest of the product is 0. In other words, the sum, as j runs from 2 to n, of T1,j×Lj,1 is 0. Remember that the top row of T equals the conjugate of the left column of L, hence their product gives a sum of squares. When a sum of squares is 0, all entries are 0. Except for the upper left entry, the top row of T, and the left column of L, are zero. This continues down the matrix, whence T is diagonal.
The matrix M is normal iff it is unitarily similar to a diagonal matrix.
Remember, we may need to extend the base field into its algebraic closure in order to build P, such that PMQ is diagonal, but it can always be done, provided M is normal.
Assume M has an eigen value v with multiplicity k. Switch to a diagonal matrix and find k orthogonal vectors corresponding to v. Independent vectors map back to independent vectors, so M has k independent eigen vectors corresponding to v.