Hermitian Conjugate of Matrix Explained

raintrek
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Simple question, and pretty sure I already know the answer - I just wanted confirmation,

Considering the Hermitian Conjugate of a matrix, I understand that

A^{+} = A where A^{+} = (A^{T})^{*}

Explicitly,

(A_{nm})^{*} = A_{mn}

Would this mean that for a matrix of A, where A is

a b
c d

that

a b
c d

=

a* c*
b* d*

=

A11 A12
A21 A22

=

A11* A21*
A12* A22*

Thanks for the clarification!
 
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And can I also ask why this seems to be a general property of the Hermitian Conjugate?

(AB)^{+} = B^{+} A^{+}

rather than

(AB)^{+} = A^{+} B^{+}
 
for your first post, you have done correct.

a b
c d

becomes

a* c*
b* d*

when you do hermitian conjugate of it.

And
(AB)^{\dagger} = B^{\dagger} A^{\dagger}

Follows from
(AB)^{T} = B^{T} A^{T}

Very easy to prove
 
As for
(AB)^{\dagger} = B^{\dagger} A^{\dagger}
and
(AB)^{T} = B^{T} A^{T}

remember that multiplication of matrices is NOT commutative.
With (AB)^{T} = B^{T} A^{T} we have (AB)^T(AB)= (A^T)(B^T B)(A)= A^T A= I. If we tried, instead, (A^TB^T)(AB) we would have (A^T)(B^T A)(B) and we can't do anything with that.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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