Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $A$?

share|cite|improve this question
up vote 4 down vote accepted

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'.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.