Question

​​​​​​​Let A be Hermitian. Show that the rank of A is equal to the number of nonzero eigenvalues of A . Is it true for non-Hermitian matrices ?

Answer 1

The rank of a Hermitian matrix is equal to the number of nonzero eigenvalues. This is not necessarily true for non-Hermitian matrices.

Step-by-step explanation:

Let A be a Hermitian matrix. The rank of A is equal to the number of nonzero eigenvalues of A.

To prove this, consider the eigenvalue equation A=, where is an eigenvector of A corresponding to the eigenvalue . Then we have A*=* where * is the conjugate transpose of .

If A is Hermitian, then A=A*. So, A*=* becomes A*=*. We can see that every nonzero eigenvalue has a corresponding eigenvector and vice versa. Therefore, the rank of A is equal to the number of nonzero eigenvalues of A.

However, for non-Hermitian matrices, it is not always true that the rank of A is equal to the number of nonzero eigenvalues. Non-Hermitian matrices can have eigenvalues with non-zero imaginary parts, which do not contribute to the rank of the matrix.