Prove that Hermitian/Skew Herm/Unitary Matrix is a Normal Matrix

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Homework Statement


Show the proof that the following are all Normal Matrices
a. Hermitian
b. Skew Hermitian
c. Unitary

Homework Equations



Normal Matrices: A*A=AA*
Hermitian Matrices: A=A* or aij=a*ji
Skew Hermitian Matrices A=-A* or aij=-a*ji

The Attempt at a Solution



So far I have tried using the above information for Hermitian Matrices to try and prove that A*A=AA* but I keep getting answers I know not to be correct. I would appreciate a nudge in the correct direction so I can quit pulling my hair out. I have a feeling when I find the correct proof it will be quite obvious, but right now I am just missing something.
 
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If A is hermitian then A*=A. A*A=AA=AA*. How are you trying to do it?
 
I was trying to do it much more in depth using amn. It never occurred to me to use the basic matrix itself. Like I said, it was right in front of me. Thank you for the 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...

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