# Thread: Matrix Solution

1. ## Matrix Solution

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.

2. Originally Posted by JoeDardeno23
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:

$c = cos \theta$ and $d = sin \theta$
where $\theta$ 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 $|b| \leq 1$.

-Dan

3. How does this relate back to the matrix though?

4. Could someone just explain this subtle point?

5. Originally Posted by JoeDardeno23
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.

-Dan