How Can We Prove the Conjugate Transpose Property of Complex Matrices?

kokolo
Messages
5
Reaction score
0
TL;DR Summary: For every Complex matrix proove that: (Y^*) * X = complex conjugate of {(X^*) * Y}

Here (Y^*) and (X^*) is equal to complex conjugate of (Y^T) and complex conjugate of (X^T) where T presents transponse of matrix
I think we need to use (A*B)^T= (B^T) * (A^T) and
Can you help me proove this cause I'm really stuck,
Thanks in advance
 
Physics news on Phys.org
kokolo said:
TL;DR Summary: For every Complex matrix proove that: (Y^*) * X = complex conjugate of {(X^*) * Y}

Here (Y^*) and (X^*) is equal to complex conjugate of (Y^T) and complex conjugate of (X^T) where T presents transponse of matrix
I think we need to use (A*B)^T= (B^T) * (A^T) and
Can you help me proove this cause I'm really stuck,
Thanks in advance
What do you know? What does ^* mean? Can you prove it for a single complex number, a ##1\times 1## matrix?

By the way: Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/
 
fresh_42 said:
What do you know? What does ^* mean? Can you prove it for a single complex number, a ##1\times 1## matrix?

By the way: Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/
## Y^* X= \overline{X^* Y}##
 
I have difficulties understanding what this is all about. Say ##\overline{X}## means the complex conjugate, ##X^T## means the transposed matrix, and ##X^\dagger=\overline{X}^T## the adjoint matrix (conjugate and transposed). Also, please write the multiplication ##X\cdot Y## with a dot. With these notations, what do you need to prove?
 
fresh_42 said:
I have difficulties understanding what this is all about. Say ##\overline{X}## means the complex conjugate, ##X^T## means the transposed matrix, and ##X^\dagger=\overline{X}^T## the adjoint matrix (conjugate and transposed). Also, please write the multiplication ##X\cdot Y## with a dot. With these notations, what do you need to prove?
##Y^* \cdot X=\overline{X^* \cdot Y}## where ##Y^*=\overline{Y^T}## and ##X^*=\overline{X^T}## and
complex conjugate matrix is ##\overline{X^* \cdot Y}##
 
You have ##(X \cdot Y)^T=Y^T\cdot X^T## and ##(X\cdot Y)^*=\overline{X\cdot Y}^T=(\overline{X}\cdot\overline{Y})^T=\overline{Y}^T\cdot \overline{X}^T=Y^*\cdot X^*.##

Does this help?
 
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...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top