# $n$ Distinct Eigenvectors for an $n\times n$ Hermitian matrix?

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?

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en.wikipedia.org/wiki/Schur_decomposition might help. –  Arkamis Nov 2 '12 at 17:54

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.