Basis of subspace (and combinations of them)


by SoapyIllusion
Tags: basis, linear algebra, subspaces
SoapyIllusion
SoapyIllusion is offline
#1
Feb13-11, 02:21 PM
P: 3
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
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vela
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Feb13-11, 03:54 PM
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Quote Quote by SoapyIllusion View Post
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]
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.
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
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.
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
spamiam
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#3
Feb13-11, 03:57 PM
P: 366
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
Robert1986 is offline
#4
Feb13-11, 03:58 PM
P: 828

Basis of subspace (and combinations of them)


Quote Quote by SoapyIllusion View Post
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
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.


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