Complex conjugate of a 5 x 5 matrices?

AI Thread Summary
To obtain the complex conjugate of a matrix, each entry is conjugated individually. The discussion clarifies that the Hermitian conjugate, or adjoint, involves taking the complex conjugate of each entry and then transposing the matrix. An example of a 5x5 matrix is provided, demonstrating the calculations for both the transpose and the Hermitian conjugate. The results for the complex conjugate and the Hermitian conjugate are shown explicitly. This highlights the distinction between the two operations in matrix algebra.
ayalam
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How do you do it?
 
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To get the complex conjugate of any matrix you just conjugate each entry.
 
Sorry I am doing something similar; converting a ket into a bra. wouldn't something change like order.
 
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matt grime is completely right for "complex conjugate". However, you may be intending "Hermitian conjugate" or "adjoint". To find the Hermitian conjugate (adjoint) of a complex matrix you take the complex conjugate of each entry and take the transpose: rows become columns.
 
Can i get a 5*5 matrices example
 
A = \left(\begin{array}{ccccc}0 &amp; i &amp; 3 &amp; 2 &amp; 1 - 2i \\ -i &amp; 7 &amp; 6 &amp; 4 &amp; 0 \\<br /> 3 + i &amp; 2 &amp; 0 &amp; -2i &amp; 4 \\ 9 &amp; 1 &amp; i &amp; 8 &amp; 1 \\ 0 &amp; 0 &amp; 0 &amp; 7i &amp; 0\end{array}\right) A^T = \left(\begin{array}{ccccc}0 &amp; -i &amp; 3 + i &amp; 9 &amp; 0 \\ i &amp; 7 &amp; 2 &amp; 1 &amp; 0 \\ 3 &amp; 6 &amp; 0 &amp; i &amp; 0 \\ 2 &amp; 4 &amp; -2i &amp; 8 &amp; 7i \\ 1 - 2i &amp; 0 &amp; 4 &amp; 1 &amp; 0\end{array}\right) A^\dagger = \left(\begin{array}{ccccc}0 &amp; i &amp; 3 - i &amp; 9 &amp; 0 \\ -i &amp; 7 &amp; 2 &amp; 1 &amp; 0 \\ 3 &amp; 6 &amp; 0 &amp; -i &amp; 0 \\ 2 &amp; 4 &amp; 2i &amp; 8 &amp; -7i \\ 1 + 2i &amp; 0 &amp; 4 &amp; 1 &amp; 0\end{array}\right)

I hope that's right and I've not made any mistakes.
 
In Nylex's example, A^T is the transpose and A^{\dagger} is the Hermitian conjugate or adjoint.
Using the same example, the "complex conjugate" that was originally asked for, and matt grimes described, would be
A^* = \left(\begin{array}{ccccc}0 &amp; -i &amp; 3 &amp; 2 &amp; 1 + 2i \\ i &amp; 7 &amp; 6 &amp; 4 &amp; 0 \\3 - i &amp; 2 &amp; 0 &amp; 2i &amp; 4 \\ 9 &amp; 1 &amp; -i &amp; 8 &amp; 1 \\ 0 &amp; 0 &amp; 0 &amp; -7i &amp; 0\end{array}\right)
 
nice examples
 
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