Recent content by Warpenguin

Well if x and y are vectors in U + W then x+y will be a vector in U + W. x + y = u + w + u' + w' = (u + u') + (w + w')? u + u' is in U and w + w' is in W. Therefore x + y is in U + W. So U + W is closed under addition. Is this right?

Char.Limit can you check if my post after that is correct?

Homework Statement Let U and W be subspaces of a vector space V Show that the set U + W = {v ∈ V : v = u + w, where u ∈ U and w ∈ W} is a subspace of V Homework Equations The Attempt at a Solution I understand from this that u and w are both vectors in a vector space V and that u+w...

double post O.o

I think what I did was wrong. If I follow what HallsofIvy said; is this how I'm supposed to do it: (x, 3z-4u-v, z, u, v)= (x, 0, 0, 0)+ (0, 3z, z, 0, 0)+ (0, -4u, 0, u, 0)+ (0, -v, 0, 0, v) = x(1, 0, 0, 0) + z(0, 3, 1, 0, 0) + u(0, -4, 0, 1, 0) + v(0, -1, 0, 0, 1) i.e...

Well I got the matrix (1 2 -1 1 2 0) (1 1 2 -3 1 0)Then I used Gaussian elimination and I'm not sure if I did it correctly but I did end up with (1 0) (0 1) which is my basis and the dimension is 2?

I'm moving on to my next section of work and i come across this example: Consider the homogeneous system x + 2y − z + u + 2v = 0 x + y + 2z − 3u + v = 0 It asks for a basis to be found for the solution space S of this system. And also what is the dimension of S. I know this might be...

Awesome! Thanks so much! I understand this much better than i did several hours ago :D

Hmm like this? ka + kc = 0 k(a + c) = 0 k(0) = 0 kb + kd + kf = 0 k(b + d + f) = 0 k(0) = 0

Yes because (a1+a2)=0, (c1+c2)=0, (b1+b2)=0, (d1+d2)=0, (f1+f2)=0 0 + 0 = 0 and 0+0+0=0

Yes that's what I meant to say xD u+v=[a1+a2 b1+b2 c1+c2] [d1+d2 e1+e2 f1+f2]

Thats what i meant when i said i wasn't sure lol. So would it be correct if i wrote u and v as 3x2 matrices containing all the elements of T and writing them as a1+a b1 +b2 etc.. what would be the next step? can i still use this after: a1+a2+d1+d2=0 (a1+a2)+(d1+d2)=0 0 + 0 = 0 or must i use all...

I'm not sure if this is the way to do it but I would say u=[a1] [d1] v=[a2] [d2] i.e if u and v are vectors in T, then u+v will be a vector in T. u+v=[a1+a2] [d1+d2] a1+a2+d1+d2=0 (a1+a2)+(d1+d2)=0 0 + 0 = 0 I'm not sure if it can be written like that

Well i'd say that it has a zero matrix because a + c = 0 (which can be 0 + 0 = 0) and b + d + f = 0 (i.e 0 + 0 + 0 =0) i.e a and d = 0. Is this correct?

Hi there. I started learning about subspaces in linear algebra and I came across a question which I'm unsure how to solve. I understand that there are 'rules' which need to be passed in order for something to be a subspace, but I have no idea how to start with this problem: Consider the set...