Question

Basic Linear Algebra: Concerning the sum of subspaces.

In my book is defined:

Answer 1

K-vectorspace V is the sum of subspaces \[W_1 + W_2 + \dots + W_m\]
if every element \[v \in V\] can be written as \[v = \sum^m_{i=1}{w_i}\], with \[w_i \in W_i\]

Now, as an example, the book proposes:
\[W_1 = \left\{ (1,0,0), (0,1,0) \right\}, W_2 = \left\{ (1,2,3), (2,3,4) \right\}, W_3 = \left\{ (1,1,1) \right\}\] It says that \[R^3 = W_1 + W_2\]

I don't understand, it's supposed to cover entire R^3, but there's only four possible combinations?

Answer 2

The main idea is that any vecror of R^3 can be represented as a sum of vectors belonging to W_1 or W_2. W_1 is actualy a canonical base for R^2. So it is just missing a vector which has non equal to zero his 3º component. (2,3,4) for example. With this 3 vectors (2 from W_1 and any of W_2) you can compose any vector in R^3

Answer 3

@xcrypt

Answer 4

You say you can compose any vector in R^3, then tell me how would you compose (5,5,5) with W_1 + W_2

Answer 5

Take any vector from W_1 and add that to any vector of W_2, you can't get (5,5,5)

Answer 6

solve linear system of equations:
x(1,0,0) + y(0,1,0)+z(2,3,4)=(5,5,5)
which is:
1x + 0y +2z = 5
0x +1y +3z =5
0x +0 y + 4z =5
_____________________
so z=5/4
y=5-(15/4)=5/4
x=5-(5/2)=5/2
so:
5/2(1,0,0) + 5/4*(0,1,0)+5/4(2,3,4)=(5,5,5)

Answer 7

The definition doesn't state that you can take linear combinations, there's no scalar \[\lambda \in K\] in there

Answer 8

And that's the whole reason why I am confused.

Answer 9

What you explained is how to do it with linear combinations of a basis of a vectorspace, not by a sum of subpaces

Answer 10

a subspace consists of all linear combinations of the basis for that subspace

Answer 11

"Linear combination" is a part of a definition of the vector space and subspace. Without that it has no sence. My solution is totaly complyant with your definition. Any vector that I used to get (5,5,5) belong to W_1 or W_2

Answer 12

Ahh, now I get it. Thanks!

Answer 13

:)