# Math Help - Direct sums and bases

1. ## Direct sums and bases

Hi, if anyone could either confirm whether my answers to these questions are right, or if not then help me find the correct answer, I'd be very grateful.

U = {(0,0,a,b) : a,b in R }
V = {(x,y,z,w) : x=2z, x+y+w=0 }
W = {(x,y,z,w) : x+y+w=0 }

Question (a): Find bases for U, V, W.

I found bases of U to be (0,0,1,0) and (0,0,0,1) - which, I hope, is simple enough.

For V I rearranged to { a(2,-1,1,-1) : a in R } following the conditions set, so does this make (2,-1,1,-1) a basis?

Similarly for W, i rearranged to { a(2,-1,0,-1) + b(0,0,1,0) : a,b in R} since z isn't specified as relating to x, y, or w in the conditions as in V. So this makes (2,-1,0,-1) and (0,0,1,0) a basis?

Question (b): Which of the sums U+V, U+W, V+W are direct?

For the sums to be direct I know that the union of the two subspaces must be {0}, but how do I prove that? Am I right in thinking that since V is a subspace of W the sum is not direct?

Question (c): Which of the sums above equal R4?

Is this when the dimension of the sums equals 4? So I have dim(U)=2, dim(V)=1 and dim(W)=2.

Using the formula dim(U+W) = dim(U) + dim(W) - dim(UnW) , so if the sum is direct I can ignore the dim(UnW) part?

Any help would be much appreciated, thanks.

2. Q. (a):

$U = sp \left\{ (0,0,1,0) , (0,0,0,1)\right\}$ , so your base for U is correct.

For V,W, the bases you provided are wrong; consider this: What is the image of $(0,2,0,-2)$ ? (There is none, so these bases are wrong!)

The correct answers would be:

$V = sp \left\{(2, -2, 1, 0) , (0, -1, 0, 1)\right\}$

$W = sp\left\{(-1, 1, 0, 0) , (0, 0, 1, 0), (-1, 0, 0, 1)\right\}$

Can you see why these are the answers? I'll expand if you want me to.

Q. (b):
U+W is not direct as $(0,0,1,0) \in U$ and $(0,0,1,0) \in W$

V+W is not direct as $(-2,2,1,0) \in V$ and $(-2,2,1,0) \in W$

V+U is a direct sum. Let us assume there exists a non-zero shared element $(x,y,z,w)$ between them, then:
$x = 2z$ so, assume $x = 0$ , then $z=0$ , then $y=-w$ but then $y \neq 0$ and then $(x,y,z,w) \notin U$

Q. (c): $dim(U) = 2$ , $dim(V) = 2$ , $dim(W) = 3$

Yes, it is correct that if $dim(A + B) = 4$ and A,B are subspaces of $\mathbb{R}^4$ , then $A + B = \mathbb{R}^4$

Can you follow from here on which two spaces give the answer?

Spoiler:
$dim(V + U) = dim(V) + dim(U) - dim(V \cap U) = 2 + 2 - 0 = 4 \Rightarrow V + U = \mathbb{R}^4$

3. Thanks for your reply Defunkt, a great help. I understand the direct sums a lot better now, but I still don't see how you got the bases for V and W. Could you please explain?

4. Also for your first answer to part (b), I can see how (0,0,-1,1) is in U but how is it in W? Since here x+y+w = 0+0+1 = 1 and not 0?

5. Oops, you're right! I meant $(0,0,1,0)$ :P

As for the bases:

$V = \left\{(x,y,z,w) \in \mathbb{R}^4 : x = 2z ; x + y + w = 0\right\} = \left\{(2z,y,z,w) : y = - 2z - w\right\}$ ... $= \left\{(2z,-2z-w,z,w)\right\} = \left\{z(2,-2,1,0) + w(0,-1,0,1)\right\}$

$W = \left\{(x,y,z,w) \in \mathbb{R}^4 : x + y + w = 0\right\} = \left\{(x,y,z,w) : x = -y - w\right\}$ ... $= \left\{(-y-w,y,z,w) \in \mathbb{R}^4\right\} = \left\{ y(-1, 1, 0, 0) + w(-1, 0, 0, 1) + z(0,0,1,0)\right\}$

Therefore:

$V = sp\left\{(2,-2,1,0), (0,-1,0,1)\right\}$

$W = sp\left\{(-1,1,0,0) , (-1,0,0,1) , (0,0,1,0)\right\}$