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I know the Gram matrix $G$ of euclidean space my operator is in and the matrix $A$ of it. How do I use them to calculate adjoined operator? I know I can use $A^*=\overline{G^{-1}A^TG}$ So for example if

$G= \begin{pmatrix} 1 & 1 & 1 & 1\\\ 1 & 2 & 2 & 2 \\\ 1 & 2 & 3 & 3 \\\ 1 & 2 & 3 & 4\end{pmatrix}$

and

$A= \begin{pmatrix} 0 & 1 & 0 & 0\\\ 1 & 0 & 1 & 0 \\\ 1 & -1 & 0 & 1 \\\ 0& 0 & 0 & 1\end{pmatrix}$

i get

$G^{-1}A^TG = \begin{pmatrix} 0 & 1 & 0 & 0\\\ 2 & 3 & 4 & 3 \\\ -1 & -3 & -3 & -2 \\\ 0& 1 &1 & 1\end{pmatrix}$

but how do I get $\overline{G^{-1}A^TG}$?

Alexander Kraynov
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  • Do you mean $G=(\langle e_i, e_j\rangle)_{i, j} $? Then I'm not sure it's $G^{-1}$ in the formula. – Berci Apr 10 '18 at 20:18
  • It is not clear what you search. If you want the Hermitian adjoint od conjugate transpose, this is simply the transpose for a matrix with real entries. If you want the adjugate matrix, than it is the transpose of the cofactor matrix. – Emilio Novati Apr 10 '18 at 20:20
  • I want to find $A^$ so that $(Ax,y)=(x,A^y)$ – Alexander Kraynov Apr 10 '18 at 20:26
  • Haven’t you’ve answered your own question? “I know I can use ...” – amd Apr 10 '18 at 22:42

1 Answers1

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Since the matrix ${G^{-1}A^TG}$ has real entries, its conjugate is the same of the matrix:

$\overline{G^{-1}A^TG}={G^{-1}A^TG}$

Emilio Novati
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