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I know that we can say if $A$ is real then it has real eigenvalues, but is the opposite always true (i.e. if $A$ has real eigenvalues then it is Hermitian)?

Quantum spaghettification
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    Look at triangular matrices. – Daniel Fischer Nov 24 '15 at 12:48
  • I would be surprised if simply looking up Hermitian matrices online would not suffice to answer this question. Anyway (beyond the non diagonalisable case) Hermitian matrices always admit an orthogonal basis of eigenvectors. So real eigenvalues but eigenspaces that are not orthogonal implies non Hermitian. – Marc van Leeuwen Nov 24 '15 at 13:07
  • By the way, what do you mean when you say $A$ is real? Note that a matrix with real entries might still have complex eigenvalues. For instance, the matrix $$ \begin{pmatrix} 0 & 1 \ -1 & 0 \end{pmatrix} $$ has characteristic polynomial $x^2 + 1$, which has no real roots. – Josse van Dobben de Bruyn Nov 24 '15 at 13:26

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The classic: $$ \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$ has one eigenvalue: $0$. It is clearly not Hermitian.

Chappers
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Not really, has to be symmetric also. For example this is not an hermitian matrix: \begin{equation} A=\left(\begin{array}{ccc} 1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right) \end{equation}

Dac0
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As pointed out by others, your statement is not true in general. It is however true for normal matrices: a normal matrix is self-adjoint if and only if it has real eigenvalues.

Community
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Josse van Dobben de Bruyn
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