Let An×n be Hermitian with eigenvalues λ1>λ2>…>λr=0 and multiplicities q1,...,qr. Can A be diagonalized? Is the matrix of eigenvalues
Ln×n=diag(λ1,…,λ1,λ2,…,λr−1,λr,…,λr)
a similar matrix to A?
Answer
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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