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Let $A_{n\times n}$ be Hermitian with eigenvalues $\lambda_1 > \lambda_2 > \ldots > \lambda_r=0$ and multiplicities $q_1,...,qr$. Can $A$ be diagonalized? Is the matrix of eigenvalues

$$L_{n\times n}=\text{diag}(\lambda_1,\ldots,\lambda_1,\lambda_2,\ldots,\lambda_{r-1},\lambda_r,\ldots,\lambda_r)$$

a similar matrix to $A$?

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Every Hermitian matrix is diagonalizable by the spectral theorem, with its eigenvalues along the diagonal, so the answer to both of your questions is `yes'.

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As you can argue by Spectral Theorem, hermitian matrices are always diagonalizable. Thus the diagonalized matrix has exactly eigenvalues along the diagonal.

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