# Linear independence of eigenvectors

1. May 4, 2009

### AxiomOfChoice

How can you show that an arbitrary $n \times n$ matrix has $n$ linearly independent eigenvectors? What if all you know about the matrix is that it's the product of a positive-definite matrix and a semi-positive-definite matrix?

2. May 4, 2009

### jbunniii

It is NOT true in general that the eigenvectors are linearly independent. It's only true for those eigenvectors corresponding to DISTINCT eigenvalues.

In the case of repeated eigenvalues, it may or may not be possible to find independent eigenvectors. For example, the identity matrix has only one eigenvalue, 1, repeated n times. In this case, EVERY nonzero vector is an eigenvector, and if you simply choose n of them at random, most likely they won't be linearly independent. (However, in this case at least it's POSSIBLE to choose n independent eigenvectors.)

For some matrices with repeated eigenvectors, it's not even possible to choose n independent eigenvectors: for example, the matrix

[0 1]
[0 0]

has only one (repeated) eigenvalue, 0. Any eigenvector [x y]^T must satisfy y = 0, so this restricts the eigenvectors to a one-dimensional subspace, and therefore you can't have two independent eigenvectors.

On the other hand, for a positive definite matrix, you are guaranteed to be able to find a linearly independent (even orthonormal) set of eigenvectors. This is even true for any Hermitian matrix by the spectral theorem (see, e.g., Horn and Johnson, "Matrix Analysis," theorem 2.5.6).

Since a positive-definite or positive-semidefinite matrix is by definition Hermitian, and so therefore is their product, then the answer to your second question is yes, it's true, but the proof is not so elementary.

3. May 4, 2009

### AxiomOfChoice

Is it possible for a Hermitian matrix to have a repeated eigenvalue, or several repeated eigenvalues? Or do we know that $n \times n$ Hermitian matrices will have $n$ distinct eigenvalues?