Question

Is the set { x E |R^4: (x1-x2)^2 + x3^4 = 0} a subspace?

Answer 1

I understand that in order to prove whether it is a subspace, there needs to closure due to addition and multiplication, but I'm getting confused how to prove (or disprove) those two conditions.

Answer 2

The obvious condition that is met is the 'zero must be part of the set', which I just take all four values (x1,x2,x3,x4) to be zero.

Answer 3

Yes, it is, because both summand are positive, so each one has to be zero, which means x3 = 0 and x1 = x2

Answer 4

So, if (x1,x2,x3,x4) and (y1,y2,y3, y4) are in the set, so is x + y, because from x1 = x2 and y1 = y2 follows x1+y1 = x2 + y2

Answer 5

Analogously with multiplication with a scalar

Answer 6

oh, and x3 and y3 have to be zero, so x3 + y3 is zero to

Answer 7

o

Answer 8

the multiplication part I'm still not fully sure about, but addition seems to make sense.

Answer 9

ok :) say you have (x1,x2,x3,x4) in the set, and you multiply it by a scalar k of R, so you want to know if (kx1,kx2,kx3,kx4) is still in the set:

From x3 = 0 follows k*x3 = k*0 = 0

From x1 = x2 follows k*x1 = k*x2

So both conditions are fulfilled, and the element is in the set, and the set is therefore closed under scalar multiplication :)

Answer 10

so x3 can be a value that isn't zero in some cases so long as the summation of the two terms equals zero (x1-x2)^2 + x3^2 = 0, right? And thank you so much for helping me!

Answer 11

Yeah, but the thing is x3 HAS to be 0, because because x3^4 and (x1-x2)^2 will ALWAYS be positive (as it is a square), and the only way to positive numbers add up to zero, is if they are both zero

Answer 12

That makes sense, thank you very much! :) Any chance I could post one more similar question that you could look at?

Answer 13

Sure, mate :) it will be my pleasure

Answer 14

Thanks a lot! I don't quite get James' answer very well, so that'd be awesome!