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I am just confused whether the following proof is true or not:

I have a square complex matrix $A$ which is also hermitian. I proved that the dimension of each eigenspace $E_{\lambda}$ is equal to $1$ (i.e, the geometric multiplicity of each eigenvalue is $1$).

To prove that all eigenvalues are distinct, I did the following:

Since $A$ is Hermitian, then it is Normal and therefore it is diagonalizable. For every diagonalizable matrix, the algebraic multiplicity of each eigenvalue=geometric multiplicity of each eigenvalue (which is equal to $1$ in this problem). Therefore, since the algebraic multiplicity of each eigenvalue is $1$, then all eigenvalues are distinct

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Your proof is correct: for Hermitian matrices you use diagonalizability to obtain the conclusion.

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