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If A is a Hermitian matrix of order n-cross-n. Then show that the rank of A is equal to the number of nonzero eigen values of A,but this is not generally true for non- Hermitian matrices.

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Since $A$ is a Hermitian matrix, it is self-adjoint, and hence normal. By the spectral theorem for normal operators, there is an orthogonal basis for $\mathbb{R}^n$ consisting of eigenvectors for $A$ which diagonalizes $A$. It should now be clear that the rank of $A$ is precisely the number of nonzero eigenvalues along the diagonal of the diagonalized matrix.

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Thank you so much.. –  Sujeet Apr 23 '13 at 16:12

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