Basis of subspace (and combinations of them)

SoapyIllusion

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 upside-down 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 upside-down 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

vela

The basis vectors must reside within the vector space you're talking about. None of those vectors is in U or V, so they can't be part of a basis for either subspace.
That's correct. The set U ∩ V is called the intersection of U and V. It's the collection of vectors common to both U and V. The set U ∪ V is called the union of U and V. It's the collection of vectors in U, V, or both sets.

spamiam

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?

Robert1986

So, if x is in U then x_1 + 2 x_2 = x_3.
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*2-1=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.