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Find if 3 vectors form a basis in space R^4

  1. May 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Is the system of vectors a1=(1,-1,0,1), a2=(2,3,-1,0), a3=(4,1,-1,4) linearly independent? Do these vectors form a basis in the vector space R^4? State why.

    2. Relevant equations



    3. The attempt at a solution

    I have done the first part of the exercise. I have found that the system of vectors is linearly independent and the rank is 3. I think that if there are 3 linearly independent vectors, and the rank is 3, then the vectors will form a basis in vector space R^3. What about R^4? How to prove whether it does form the basis?
     
  2. jcsd
  3. May 4, 2012 #2
    Try to see if they could span R4. The answer is no, they won't span it. You need 4 vectors for a basis in R4 because the dimension is 4, but that's a theorem, try to figure out why there must be a vector that is not a linear combination of these 3 vectors. (Looking at the system of linear equations will give you some insight into it).
     
  4. May 4, 2012 #3

    Mark44

    Staff: Mentor

    No, they can't possibly form a basis for R3. They are vectors in R4. They form a basis for a three-dimension subspace of R4, but that is very different from R3.
    The vectors can't form a basis for R4 since there aren't enough of them.
     
  5. May 4, 2012 #4

    sharks

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    I never quite understood the R^n or R^m concept. I tried but gave up. What do you mean when you say "since there aren't enough of them". Is R^n representative of the total number of rows or columns or unique number of independent vectors?
     
  6. May 4, 2012 #5

    Mark44

    Staff: Mentor

    Vectors in R or R1 have one component (a single real number). Any basis for this vector space contains one vector.
    Vectors in R2 have two components (e.g., <1, 3>). Any basis for this vector space contains two vectors.
    Vectors in R3 have three components (e.g., <1, 3, -2>). Any basis for this vector space contains three vectors.
    And so on...

    In the problem in this thread we're given three vectors in R4. Being vectors in R4, they have four components, so they don't belong to, for example, R3 or R2 or R5. A basis for R4 must be a linearly independent set of four vectors.
     
  7. May 4, 2012 #6

    sharks

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    OK, as an example, if a 4x5 matrix has only 2 linearly independent vectors, then these 4 row vectors form a basis in R^2, corresponding to the number of pivot rows or pivot columns only. Correct?
     
  8. May 4, 2012 #7

    Mark44

    Staff: Mentor

    No.

    A 4 x 5 matrix is a map from R5 to R4. The rows are vectors in R5, and have absolutely nothing to do with R2.
     
  9. May 4, 2012 #8

    sharks

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    I'm sorry, but i don't understand. What about the effect of the number of independent linear vectors on R^n or R^m?

    Is there a good book or paper that i could refer to? I just can't wrap my head around those R^n or R^m numbers. But i do know the fundamental theorem of linear algebra part 1, which deals with dimensions.

    EDIT: OK, i think i understand. The R^{value} is always equal to the number of components in the row/s corresponding to the linearly independent vector/s. So, the rule is always count the number of components in the rows and not in the independent columns. Correct?
     
    Last edited: May 4, 2012
  10. May 4, 2012 #9

    Mark44

    Staff: Mentor

    Now I don't understand what you're asking.

    Let's go back to your example of a 4 x 5 matrix.

    Suppose you ended up with these vectors:
    <1, 0, 1, 1, 1>
    <0, 1, 1, 1, 1>
    You asked whether these could be a basis for R2. They couldn't possibly be, because these two vectors belong to R5 (they have 5 components). A basis for R2 would have to consist of vectors in R2, such as <1, 1> and <0, 1>.

    The two vectors in R5 above are a basis for a two-dimensional subspace of R5. This subspace looks like a plane, but it's a plane embedded in 5-dimensional space. The fact that this subspace of R5 is a plane has nothing to do with the R2 vector space.
    It's just a matter of realizing that n and m in the symbols Rn and Rm refer to vector spaces of dimension n and m respectively. n and m are just placeholders for some integers.

    You aren't going to be able to come up with a mental image of 5-dimensional space (or 4- or 6- or n-) but that's OK - you don't need to. You can visualize a 1-dimensional subspace as a line in some blob that represents the higher-dimension space. A 2-dimensional subspace is just a plane in the higher-dimension space.

    What is it about Rn that you're having trouble with?
     
  11. May 4, 2012 #10

    Mark44

    Staff: Mentor

    You made your edit after I had already started a response.
    That's more complicated than it needs to be. The "value" is equal to the number of components in a vector in R{value}.
    In your example of a 4 x 5 matrix that row-reduces to a matrix with two non-zero rows, Rank(A) = 2, Kernel(A) = 3, and n = 5, where n is the dimension of the domain, R5.
     
  12. May 4, 2012 #11

    sharks

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    That's the best way that i can understand and remember it. It's just the rank-nullity theorem.
     
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