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I have the matrix H:
A11=a+bi
A12=c+di
A21=-c+di
A22=a-bi
I need to show that the equation x^2=-1 has infinitely many solutions in H.
I ignored a (because the book told me to) and so I have the matrix:
A11=bi
A12=c+di
A21=-c+di
A22=-bi
where b^2+c^2+d^2=1.
I'm just having trouble showing the "infinitely many solutions in H" part.
You have one constraint on three variables. So, for example, given a value b = 0, how many solutions can you come up with for c and d? I get an infinite number:Quote:
Originally Posted by JoeDardeno23
$\displaystyle c = cos \theta$ and $\displaystyle d = sin \theta$
where $\displaystyle \theta$ is a continuous parameter.
This alone gives you an infinite number of solutions, but you can obviously make a similar kind of argument for any acceptable value of b, that is for $\displaystyle |b| \leq 1$.
-Dan
How does this relate back to the matrix though?
Could someone just explain this subtle point?
Your matrix is defined by the values b, c, and d by the A11, A12, A21, and A22 relations that you gave in your initial post. The condition that any matrix belonging to the set H be such that x^2 = -1 gives a constraint on the possibilities that b, c, and d can take, but is not so resistrictive as to give a finite set of possibilities.Quote:
Originally Posted by JoeDardeno23
-Dan
