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Difference between span and basis

  1. Dec 12, 2014 #1
    I'm just having a small trouble understanding the difference ( occurred while I was doing exercise).

    A basis is defined as
    1)linearly independent
    2)spans the space it is found in.

    Here is where I get confused:

    To determine whether or not a set spans a vector space, I was taught to find its determinant and if det|A|=/= 0 then it spans the space.
    I was also taught that if det|a|=/=0 then it isn't coplanar and therefore it is linearly independent ( can also just solve to see if trivial sol'n...)

    But then if both det|a=/= it means that it spans and is linearly independent. Therefore, in my head, it comes with the idea that "spans is related to independence"

    Anybody got a good way to differentiate both?
     
  2. jcsd
  3. Dec 12, 2014 #2

    Stephen Tashi

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    How were you taught to form the matrix whose determinant you take?
     
  4. Dec 12, 2014 #3
    lets say its in r^3, multiply the 3 vectors by a coefficient k1,k2,k3. put the coefficient in a matrix, find the determinant.
     
  5. Dec 12, 2014 #4

    Mark44

    Staff: Mentor

    A basis is a set of vectors that is
    1)linearly independent
    2)spans the space or subspace it is found in.
    What does "it isn't coplanar" mean? Who is "it"?
    Not necessarily. This set spans R3 but isn't linearly independent.
    {<1, 0, 0>, <0, 1, 0>, <0, 0, 1>, <1, 1, 1>}

    This set is linearly independent, but doesn't span R3.
    {<1, 0, 0>, <0, 1, 0>}

    A spanning set for a space or subspace is a set of vectors for which every vector in the space/subspace is a linear combination of the vectors in the spanning set. In my first example above, every vector in R3 can be written as a linear combination of the four vectors in the set.

    A set of n vectors {v1, v2, ... , vn} is linearly independent if the only solution to the equation c1 v1 + c2 v2 + ... + cn vn = 0 is the trivial solution (i.e., c1 = c2 = ... = cn = 0).

    A basis for a space/subspace is a set of vectors that spans the space/subspace and is a linearly independent set. If the dimension of the space or subspace is n, a spanning set must have at least n vectors in it. A linearly independent set can have at most n vectors in it. A basis is a minimal spanning set.
     
    Last edited: Dec 12, 2014
  6. Dec 12, 2014 #5

    Stephen Tashi

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    Can you give a web link to an explanation of this method?

    I don't know what you mean by "multiply the 3 vectors by coefficient k1,k2,k3". Do you have the unknowns k1,k2,k3 in the matrix?

    If you put the 3 vectors (as n-tuples of given numbers in R^3) in the matrix then you have a square matrix and can take the determinant. But what does the method tell you to do if you have less than 3 vectors ? What does it say to do if you have more than 3 vectors?
     
  7. Dec 16, 2014 #6

    Fredrik

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    I don't think this is an improvement over the OP's definition. The set ##\{(1,0,0)\}## "is found in" (I can only assume that this means "is a subset of") infinitely many subspaces of ##\mathbb R^3##, but it only spans one of them. If you meant that the set spans the space it spans, then the second statement isn't saying anything.

    The OP's definition is fine in my opinion. If I had to change something about it, I would add some clarity by mentioning the space for which the set is supposed to be a basis:

    Let ##V## be a vector space. A set ##B\subseteq V## is said to be a basis for ##V## if
    (a) ##B## is linearly indendent.
    (b) ##B## spans ##V##.

    Let ##V## be a vector space. Let ##S## be a non-empty subset of ##V##. Let ##W## be a subspace of ##V##. The following statements are equivalent:

    (a) ##W## is the intersection of all subspaces of ##V## that has ##S## as a subset.
    (b) If ##U## is a subspace of ##V## such that ##S\subseteq U##, then ##W\subseteq U##. (In other words, ##W## is the smallest subspace that contains ##S##).

    For each non-empty subset ##S##, we define the span of ##S## as the unique subspace ##W## that satisfies the equivalent conditions above. Different books use different notations for this subspace. Some common ones are ##\operatorname{span} S## and ##\bigvee S##. The set ##S## is said to span ##W##, to generate W, and to be a spanning set for ##W##. (These statements all mean the same thing, that the equivalent conditions above are satisfied).

    Let ##V## be a vector space. Let ##S## be a subset of ##V##. The following statements are equivalent.

    (a) ##S## is a maximal linearly independent set in ##V##.
    (b) ##S## is a minimal spanning set for ##V##.

    If you're not familiar with the minimal/maximal terminology, then these statements need to be explained. This is what they mean:

    (a) For all ##T\subseteq V##, if ##T## is linearly independent, then ##T\subseteq S##.
    (b) For all ##T\subseteq V##, if ##T## spans ##V##, then ##S\subseteq T##.

    One book I read used this theorem to define the term "basis". Such a definition would look like this:

    Let ##V## be a vector space. A subset ##S\subseteq V## is said to be a basis for ##V## if it satisfies the equivalent conditions of the theorem. (In other words...if it's a minimal spanning set for ##V##, or equivalently, a maximal linearly independent set in ##V##).

    This approach can't be applied to sets like ##\{(1,0,0),(0,1,0)\}\subseteq R^2##, because the matrix ##\begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\end{pmatrix}## isn't even square. To get a square matrix when you consider a subset of ##\mathbb R^n##, you have to start with a set with ##n## elements. If you find that the determinant is non-zero, this tells you that you have found a linearly independent set. Since it has ##n## elements, and ##\mathbb R^n## doesn't contain any linearly independent subset with ##n+1## elements, it's a maximal linearly independent set, and therefore a basis.
     
  8. Dec 16, 2014 #7

    Mark44

    Staff: Mentor

    All I was doing was to make what the OP (MarcL) wrote somewhat clearer. For instance, "A basis is defined as 1)linearly independent" omits the idea that we're talking about a set of vectors. Much of his wording I left the same. If I had written my own definitions, I would have used a somewhat different wording.
     
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