Is S Closed Under Addition in R^(2x2)?

  • Thread starter Thread starter WTFsandwich
  • Start date Start date
WTFsandwich
Messages
7
Reaction score
0

Homework Statement


Suppose A is a vector \in R^{2x2}.

Find whether the following set is a subspace of R^{2x2}.

S_{1} = {B \in R^{2x2} | AB = BA}


The Attempt at a Solution


I know that S isn't empty, because the 2 x 2 Identity matrix is contained in S.

The problem I'm having comes in the proof that addition is closed.

If I show A(B + C) = (B + C)A that should be sufficient, right?

So far I have:

Suppose B and C \in S.
A(B + C) = (B + C)A
AB + AC = BA + CA

And that's where I'm stuck. I have no idea where to continue on to. Any help would be greatly appreciated.
 
Physics news on Phys.org
You need to show that S is closed under addition and scalar multiplication. You probably want to do those separately.
 
I know that's what I have to do, but I don't know how to go about doing it. I started the addition part up above, and am stuck at that point.
 
OK, B and C are both elements of S.
A(B + C) = AB + AC (since vector multiplication is left-distributive)
Now, what can you say about AB and AC, since B and C are members of set S?
 
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...
Back
Top