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Much like the title says, I wish to know how it is possible that we can know that there are $n$ distinct eigenvectors for an $n\times n$ Hermitian matrix, even though we have multiple eigenvalues. My professor hinted at using the concept of unitary transform and Gram-Schmidt orthogonalization process, but to be honest I'm a bit in the dark. Could anyone help me?

share|cite|improve this question might help. – Emily Nov 2 '12 at 17:54
up vote 1 down vote accepted

You can show that any matrix is unitarily similar to an upper triangular matrix over the complex numbers. This is the Schur decomposition which Ed Gorcenski linked to. Given this transformation, let $A$ be a Hermitian matrix. Then there exists unitary matrix $U$ and upper-triangular matrix $T$ such that $$A = UTU^{\dagger}$$ We can show that any such decomposition leads to $T$ being diagonal so that $U$ not only triangularizes $A$ but in fact diagonalizes it.

Since $A$ is Hermitian, we have $$A= UT^{\dagger}U^{\dagger} = UTU^{\dagger} = A^{\dagger}$$ This immediately implies $T^{\dagger} = T$. Since $T$ is upper-triangular and $T^{\dagger}$ is lower-triangular, both must be diagonal matrices (this further shows that the eigenvalues are real). This shows that any Hermitian matrix is diagonalizable, i.e. any $n\times n$ Hermitian matrix has $n$ linearly independent eigenvectors.

share|cite|improve this answer
I see. I understand this proof, but the what was my professor talking about with the Gram-Schmidt orthogonalization process? I think this Shur process is a form of unitary transform, but it has nothing to do with the other thing my professor suggested. – Additional Pylons Nov 3 '12 at 1:44
Well, the Gram-Schmidt procedure is a step in the proof of the Schur decomposition but I really don't know. – EuYu Nov 3 '12 at 1:48
I see, but a few points I would like to ask is, since A is Hermitian, wouldn't that make A equal to its conjugate transform, not just its transform? And I also feel that this is just a more specific case of diagonalization. Am I right or am I just not seeing something? – Additional Pylons Nov 3 '12 at 6:58
This is a more specific form of diagonalization. It's call unitary diagonalization. And what do you mean by "equal to its conjugate transform, not just its transform?" – EuYu Nov 3 '12 at 7:08

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