There are advantages to working with complex numbers.

If is a complex matrix, the conjugate-transpose of is defined entrywise by For complex matrices, this is the analogue of the transpose. A complex matrix is said to be unitary if , and Hermitian if (these are the complex analogues of orthogonal and symmetric, respectively). The proof of the following theorem involves definitions and methods developed in the next module; it is included here for completeness. The theorem will allow us to determine necessary and sufficient conditions for diagonalizability.

Proof
By the Fundamental Theorem of Algebra factors completely over into linear terms. In particular, must have at least one eigenvalue. Let be an eigenvalue of , and a corresponding eigenvector with norm 1. Let be the orthogonal complement of the span of . Write for the linear transformation . Now consider the linear transformation By induction on dimension, we can assume the subspace admits a complex orthonormal basis such that . In other words, for each we have This means that the composition on the vectors is given by for some choice of complex scalars . Setting , we see that Let be the concatenation of the vectors and the upper triangular matrix with . Then is unitary, and last equation can then be written in matrix form as Multiplying both sides on the left by gives the desired equation

The main corollary to Shur’s theorem is

Proof
By Shur’s Theorem, there exists a unitary and upper triangular with . But is Hermitian, so . Then

But as is upper triangular, the identity implies not only that is diagonal, but that the diagonal elements remain invariant under complex conjugation; i.e., are real numbers. Thus Shur’s equation may be rewritten in this case as where is a real diagonal matrix, or alternatively, But this last equation implies , where . As the columns of are orthonormal, they form a basis for , implying the result.

As a special case, when is a real matrix, then it is Hermitian iff it is symmetric. Thus, viewing it as a complex matrix with purely real entries, we have

A Hermitian matrix is said to be positive definite if all of its eignevalues are positive real numbers, and non-negative definite, if all of its eigenvalues are non-negative. Now suppose is an arbitrary complex matrix, and . Then clearly implying is Hermitian.

Proof
Since admits an orthonormal basis consisting of eigenvectors of , it suffices to show that the eigenvalues of are non-negative. Let be an eigenvector of with eigenvalue . Then As this implies . Also, as we have already seen, is non-singular iff is not an eigenvalue of . But . Hence is non-singular iff is.