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Basis of subspace (and combinations of them) 
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#1
Feb1311, 02:21 PM

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1. The problem statement, all variables and given/known data
We are given the following subspaces U := {x E R3: x1 + 2*x2  x3 = 0} and V := {x E R3: x1  2*x2  2*x3 = 0} And we need to find a basis for (i) U (ii) V (iii) U n V (not an "n" but a symbol that looks like an upsidedown U) (iv) span(U u V) (not a "u" but a symbol that looks like a U) 2. The attempt at a solution Because x is a subspace of R3 in both V and U, it seemed that for (i) and (ii) the trivial basis would simply be e1 = [1,0,0] e2 = [0,1,0] and e3 [0,0,1] I also do not know what the U and upsidedown U symbols mean, but someone guessed that "n" meant where they overlap and "u" meant the combination of both subspaces The answers I found seem to trivial, am I missing something very obvious, and could anyone give me any suggestions to lead me in the right direction 


#2
Feb1311, 03:54 PM

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#3
Feb1311, 03:57 PM

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You need to use the relations given in the definition of these sets to find a basis. Consider this similar example. Suppose I'm given the set
[tex] S = \{(x_1, x_2) \in \mathbb{R}^2 : x_1 = x_2\} [/tex] I set the common value of [itex]x_1[/itex] and [itex]x_2[/itex] to the variable [itex] t [/itex] and I get [tex] \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}= \begin{pmatrix} t \\ t \end{pmatrix}= t \begin{pmatrix} 1 \\ 1 \end{pmatrix} [/tex] Thus the vector (1,1) forms a basis for my subspace S. Can you do something similar for your subspaces? 


#4
Feb1311, 03:58 PM

P: 828

Basis of subspace (and combinations of them)
Ok, first, ask yourself what a basis is. Now, if e1,e2 and e3 formed a basis of U, then this would imply that the vector (1,1,1) is in U. However, 1+2*21=4, not 0. Thus, e1,e2,e3 do not form a basis. So, I'm gonna help you out for (i) and then try to figure it out for the rest. (a,0,a) and (0,b,2b) form a basis for this subspace (where a and b are fixed real numbers). Why is this a basis? Well, any linear combination of these two vectors: A(a,0,a) + B(0,b,2b) = (a,b,A a + 2B b) is clearly in U. Also, if x is in U, it can be written as a linear combination of (a,0,a) and (0,b,2b). Furthermore, the two vectors are clearly independent. So, do you see how it goes? And you are correct the upside down U means intersection, that is, where they overlap, and you were also right about the U. 


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