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I have the following matrix

$$A = \pmatrix{1 & 0 & 1\cr 0 & 2 & 0\cr 1 & 0 & 1\cr}$$

and

$$adj(\bf{A})=\pmatrix{2 & 0 & -2\cr 0 & 0 & 0\cr -2 & 0 & 2\cr}$$

MATLAB says the matrix $A$ is Hermitian. However the conjugate transpose is not equal the matrix $A$. How to check the matrix is Hermitian ?

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If you mean $$A = \pmatrix{1 & 0 & 1\cr 0 & 2 & 0\cr 1 & 0 & 1\cr}$$ that is Hermitian: its conjugate transpose (which in this case is the transpose, because the matrix is real) is itself. As for $\pmatrix{2 & 0 & -2\cr 0 & 0 & 0\cr -2 & 0 & 2\cr}$, that is the "classical adjoint" or adjugate. It has nothing to do with the Hermitian adjoint or conjugate transpose. The terminology is unfortunately confusing.

The Matlab command

adjoint(A)

gives you the classical adjoint of $A$, while

A'

gives you the Hermitian adjoint.

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  • $\begingroup$ You mean that Classical Adjoint is different from Hermitian Adjoint ?? :O I did not know that. $\endgroup$
    – Bruno Gallego
    Commented Feb 18, 2021 at 13:45

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