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Help me understand what a basis is

  1. Nov 12, 2017 #1
    1. The problem statement, all variables and given/known data
    I'm using this problem, and although I got the correct answer my question is really about the properties of a basis, not the problem that I'm going to state..

    anyways,

    Let = {v_1, v_2, ..., v_n } be a basis for a vector space V. Let c be a nonzero scalar. Show that the set {cv_1, cv_2, ..., cv_n } is also a basis for V

    2. Relevant equations


    3. The attempt at a solution
    Since the set {cv_1, cv_2, ..., cv_n } has the same dimensions as set {v_1, v_2, ..., v_n }, I now only need to check for linear independence (or span, but it makes no sense to check for the more difficult condition.)

    So [ k1cv1 + k2cv2 + ... kncvn = 0

    the set consists of vectors {v_1, v_2, ..., v_n }, which was stated to be a basis, thus implying that any set with these vectors must = 0 for each constant,since c is a nonzero scalar, and this must mean that k1 through kn are all zero.

    Does this seem like a logical/correct conclusion?

    Anyways, my real question is:

    It says that c is any nonzero scalar, but this doesn't make sense to me.

    I was under the impression that a vector space can only have one basis. This problem contradicts that belief. Maybe my understanding of the terminology is wrong, but if a basis is a scalar multiple of another basis, they are essentially the same basis, right?

    Sorry I know my wording for this question doesn't really make sense, I hope you guys can understand what I'm saying.

    Through finishing this question, my train of thought is as follows:

    Since a basis is
    1) a linear combo which will = 0 if and only if the constant multiple is = 0, (its linearly independent)
    2) spans every vector in the vector space (meaning every single vector in the vector space can be formed by the basis)

    that means that you can have "multiple" basis, all being scalar multiplies of each other, because by taking a scalar multiple of those multiple basis, you can construct any vector in the vector space V.

    Again I apologize because reading this I truly understand how stupid my questions sound but I'm having a hard time grasping these concepts, so please bear with me.

    Also I posted this in precalculus since the problem doesn't involve any understanding of calculus. I hope this is in the correct section and I ask of the mods to move it if its not.
     
  2. jcsd
  3. Nov 12, 2017 #2

    FactChecker

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    Science Advisor
    Gold Member

    You are over-thinking this. Consider any vector in the space, write it in terms of the given basis vectors {v_1, v_2, ..., v_n }, convert that expression to a linear combination of {cv_1, cv_2, ..., cv_n }. That does it.

    You are wrong about there only being one set of basis vectors. Not only any nonzero multiplier, but lots of other linear combinations of a basis will also be a basis. Like {v_1, v_2+v_1, ...., v_n+v_1} will be a basis.
     
  4. Nov 13, 2017 #3

    Mark44

    Staff: Mentor

    I don't think so.
    Let's consider a simpler space, ##\mathbb R^2## and a set of two vectors: ##\vec u = <1, 0>## and ##\vec v = <2, 0>##.
    I notice that the equation ##c_1\vec u + c_2 \vec v = \vec 0## is true when ##c_1 = c_2 = 0##. Does this imply that 1) ##\vec u## and ##\vec v## are linearly independent, and 2) that these two vectors form a basis for ##\mathbb R^2##? After all, there are two vectors.
     
  5. Nov 17, 2017 #4
    Isn't that different from this cause though, because it's not explicitly stated that u and v form a basis?
     
  6. Nov 17, 2017 #5

    Mark44

    Staff: Mentor

    My point is that for any vectors u and v, we can always set up the equation ##c_1\vec u + c_2\vec v = \vec 0##, and find that ##c_1=0## and ##c_2 = 0##, whether or not u and v are linearly independent.
     
  7. Nov 17, 2017 #6

    Mark44

    Staff: Mentor

    My earlier reply was base on what you said here.
    I didn't notice that you had said this, later on.
    That "if and only if" part is a subtlety that a lot of students miss. Clearly you get it, so my earlier post wasn't needed.
     
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