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Vector subspace

  1. Feb 18, 2008 #1
    1. The problem statement, all variables and given/known data

    Are the vector subspaces U={(x,y,0,0) | x+2y=0} and W= {(0,0,z,t | z+t=0} from [tex]R^4[/tex] stand for U+W = [tex]R^4[/tex]


    2. Relevant equations



    3. The attempt at a solution

    Can somebody explain, how will solve this task. I have no idea, how they do in my book.
    Thanks.
     
  2. jcsd
  3. Feb 18, 2008 #2

    HallsofIvy

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    "Stand for"? Do you mean to ask "is the direct sum of U and W equal to R4"?

    Each of U and V has dimension one so their direct sum will have dimension 2. It cannot possible by equal to R4!
     
  4. Feb 18, 2008 #3
    Why it cannot be possible by equal to R4? Can you explain how you solved, this task?
     
  5. Feb 18, 2008 #4

    Tom Mattson

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    Think, man, think!:wink: What's the dimension of [itex]\mathbb{R}^4[/itex].
     
  6. Feb 18, 2008 #5
    But aren't there any calculations. Sorry for my English, but I can't understand what you want to say.
     
  7. Feb 18, 2008 #6

    Tom Mattson

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    OK, let's step through this. Do you understand why [itex]dim(U \oplus W)=2[/itex]?
     
  8. Feb 18, 2008 #7
    No, I don't understand it.
     
  9. Feb 18, 2008 #8

    Tom Mattson

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    OK, let's write these subspaces another way. [itex]U=span\{(-2,1,0,0)\}[/itex] and [itex]W=span\{(0,0,1,-1)\}[/itex]. Clearly, [itex]dim(U)=dim(W)=1[/itex]. Since [itex]U \cap W=\{(0,0,0,0)\}[/itex], it follows that [itex]U+W=span\{(-2,1,0,0),(0,0,1,-1)\}[/itex]. So, [itex]dim(U+W)=2[/itex].

    What we actually have here is a direct sum [itex]U \oplus W[/itex] of the subspaces. If your texbook covers direct sums, then it probably also proves that the dimension of the direct sum of two subspaces equals the sum of the dimensions of the individual subspaces. But even if it doesn't, you can clearly see that [itex]U+W[/itex] has 2 independent vectors in its basis, so it has dimension 2.

    So if you know what the dimension of [itex]\mathbb{R}^4[/itex] is, it's obvious why they can't possibly be equal.
     
  10. Feb 18, 2008 #9
    Ok, I understand know. Thank you very much. Just can you tell me what is dimension of [itex]\mathbb{R}^4[/itex]?
     
  11. Feb 18, 2008 #10

    Tom Mattson

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    If you don't know, I really think you should look it up.
     
  12. Feb 18, 2008 #11
    Ok, dim[itex]\mathbb{R}^4[/itex]=4

    And if I have U and W vector subspaces.
    U={ (x,y,z) | x+2y+3z=0 } and W= {(x,y,z) | x+y+z=0 }

    Prove that [itex]U \oplus W[/itex] = [itex]\mathbb{R}^3[/itex]

    How will you solve this problem? Thanks.
     
  13. Feb 18, 2008 #12

    Tom Mattson

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    That's right. Now it should be clear why [itex]U+W \neq \mathbb{R}^4[/itex].

    Something is wrong with this problem. You've got two 2-dimensional subspaces of [itex]\mathbb{R}^3[/itex]. Clearly, they cannot intersect in only the zero vector, so this isn't a direct sum. Perhaps you meant [itex]U+W=\mathbb{R}^3[/itex]? If so, then I would begin by writing down the basis of each subspace and putting them into a list. This will give you a list of 4 vectors, and since they are elements of [itex]\mathbb{R}^3[/itex], at most 3 of them can be linearly independent. You need to find exactly 3 that are independent.
     
  14. Feb 18, 2008 #13
    How you will find base for U or W?
     
  15. Feb 18, 2008 #14

    Tom Mattson

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    That should be clear from the definitions of U and W. Each definition explicitly tells you what is required for a vector to belong to the subspace. Haven't you seen any examples of problems like this?

    Here is a simple example.

    Find a basis for the subspace [itex]U=\{(x,y,z)|x=y\}[/itex] of [itex]\mathbb{R}^3[/itex].

    Solution:

    An arbitrary vector [itex]u\in U[/itex] can be written as:

    [tex]u=(x,x,z)[/tex]

    [tex]u=(x,x,0)+(0,0,z)[/tex]

    [tex]u=(1,1,0)x+(0,0,1)z[/tex]

    So clearly, [itex]U=span\{(1,1,0),(0,0,1)\}[/itex]. Furthermore, [itex]dim(U)=2[/itex].

    Get it?
     
  16. Feb 19, 2008 #15
    Ok, I understand now. But for my case it is little complicated.

    [itex]U=\{(x,y,z)|x+2y+3z=0\}[/itex] of [itex]\mathbb{R}^3[/itex].

    [tex]u\in U[/tex]

    [tex]u=(x,\frac{x+2y-x}{2},\frac{-x-2y}{3})[/tex]

    [tex]u=(x,y,0)+(0,0,\frac{-x-2y}{3})[/tex]

    [tex]u=(x,0,0)+(0,y,0)+(0,0,\frac{-x-2y}{3})[/tex]

    [tex]u=(1,0,0)x+(0,1,0)y+(0,0,1)\frac{-x-2y}{3}[/tex]

    [itex]U=span\{(1,0,0),(0,1,0),(0,0,1)\}[/itex]. So [itex]dim(U)=3[/itex]

    Now, for [tex]W[/tex].

    [itex]W=\{(x,y,z)|x+y+z=0\}[/itex] of [itex]\mathbb{R}^3[/itex].

    [tex]w\in W[/tex]

    [tex]w=(x,-x-z,-x-y)[/tex]

    [tex]w=(x,-x+x+y,-x-y)[/tex]

    [tex]w=(x,y,-x-y)[/tex]

    [tex]w=(x,y,0)+(0,0,-x-y)[/tex]

    [tex]w=(1,0,0)x+(0,1,0)y+(0,0,1)(-x-y)[/tex]

    [itex]W=span\{(1,0,0),(0,1,0),(0,0,1)\}[/itex]. So [itex]dim(W)=3[/itex]

    [itex]dim(W)+dim(U)=6[/itex]

    Is this correct? Thanks.
     
    Last edited: Feb 19, 2008
  17. Feb 19, 2008 #16

    HallsofIvy

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    No, it is not at all correct. How can you be expected to do problems like this and tell us that you do not know what "dimension" means and do not know what "direct sum" means? And, frankly, that is exactly what you are saying. U and W are both subsets of R3. Their direct sum is also a subset of R3 and so cannot have dimension higher than 3. Even if it were true that "dim(W)+ dim(U)= 6" but that tells you nothing about the dimension of their direct sum.

    U= {(x,y,z)| x+ 2y+ 3z= 0}. You can, for example, solve that equation for x: x= -2y- 3z. Now, if we take y= 0, z= 1, we get x= -3 so <-3, 0, 1> is a vector in U. If we take y= 1, z= 0, we get x= -2 so <-2, 1, 0> is a vector in U. In fact it is easy to show that those two vectors form a basis for U (that's the advantage to using "0" and "1" for y and z) so the dimension of U is 2, not 3.

    W= {(x,y,z)| x+ y+ z= 0}. z=-x-y so if we take x= 1, y= 0,we get z= -1. <1, 0, -1> is a vector in W. If we take x= 0,y= 1, we get z= -1 so <0, 1, -1> is a vector in W. Again, it is easy to see that they form a basis for W. W has dimension 2, not 3. But their direct sum is still a subspace of R3- the dimension of the subspace must be less than or equal to 3 so the sum, 2+ 2, is not the dimension of their direct sum.

    The direct sum combines the bases. Since {<-2, 1, 0> , <-3, 0, 1>} is a basis for U and {<0, 1, -1>, <1, 0, -1>} is a basis for W, {<-2, 1, 0>, <-3, 0, 1>, <0, 1, -1>, <1, 0, -1>} spans their direct product. But that is not a basis because the four vectors are not independent. You need to find a subset of those 4 that are independent. The number of independent vectors that span the direct product is its dimension.

    If you do not understand any of those words, like "independent" and "span", look them up in your text book. They are essential to defining "dimension" of a vector space.
     
  18. Feb 19, 2008 #17
    Ok, I understand what you want to say. Thanks. Let's say we have U and W subsets of[tex]R^3[/tex]? Do always U and W must have dim=3? Can sometimes the dimension of U+W from [tex]R^3[/tex] be 4? Do must 3 of them to be linear combination of the 4-th of them? Or the 1st+2nd vector can be linear combination of the 4th one?
     
    Last edited: Feb 19, 2008
  19. Feb 20, 2008 #18
    I have one new task.

    Find [tex]U+W[/tex] [tex]\subseteq[/tex] [tex]\mathbb{R}^3[/tex]

    [tex]U=\{(a,b,0)| a=b,a,b[/tex] [tex]\in[/tex] [tex]\mathbb{R}\}[/tex]

    [tex]W=\{(0,b,c)| b=c, b,c[/tex] [tex]\in[/tex] [tex]\mathbb{R}\}[/tex]

    Can somebody help?
     
  20. Feb 20, 2008 #19
    How do you know that it will give me a list of 4 vectors?
     
  21. Feb 20, 2008 #20

    HallsofIvy

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    If U and W are only sets and not subspaces they don't necessarily have any "dimension" at all! No, it is not true that subspaces of R3 must have dimension 3. It is true that they cannot have dimension greater than 3. As I said before, any combination of subspaces of R3 is still a subspace of R3 and cannot have dimension larger than 3. "must 3 of them to be linear combination of the 4-th of them" is the wrong way around. Since there cannot be more than 3 independent vectors in dimension 3, one of them must be a linear combination of the other 3. (I'm going to assume that you have a little problem with the English, but really understand what a "linear combination" is.) Of course, it might be that two of the vectors can be written as a linear combination the other 2 (in which case we have dimension 2) or that all are just some multiple of one (dimension 1).

    Because "a= b", U is just the set of vectors of the form (a, a, 0)= a(1, 1, 0). That is one dimensional- {(1, 1, 0)} is a basis. Because "b= c", W is just the set of vectors of the form (0, b, b)= b(0, 1, 1). That is also one dimensional with basis {(0, 1, 1)}. Clearly those two vectors are independent- the only way we could have a(1, 1, 0)+ b(0, 1, 1)= (a, a, 0)+ (0, b, b)= (a, a+ b, b)= (0, 0, 0) is to have a= b= 0- so the two {(1, 1, 0), (0, 1, 1)} form a basis for U+ W which has dimension 2.

    For each subspace you had one linear equation in x, y, z which you could solve for one in terms of the other two (say, z as a function of x and y). That means that by taking different values for x and y (I prefer x= 0, y= 1 and then x= 1, y= 0) we get two values of z, giving a basis of two vectors for each subspace. Those are the "4 vectors" Tom Mattson was referring to. Of course, they might not be independent so the set of all 4 might not form a basis for the direct sum.
     
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