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Question involving a linearly independent set of vectors

  1. Feb 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that if {a, b, c} is a linearly independent set of vectors, then so are {a, b}, {a, c}, {b, c}, {a}, {b}, and {c}.

    2. Relevant equations


    3. The attempt at a solution

    Well I was just thinking that if {a, b, c} is a linearly independent set of vectors, then {a, b, c} span R3. And so if {a, b, c} span R3, then {a, b}, {a, c}, and {b, c} must span R2 and be linearly independent, and {a}, {b}, and {c} span R1 and be linearly independent. That's all I got. How could I prove this mathematically?

  2. jcsd
  3. Feb 29, 2012 #2
  4. Feb 29, 2012 #3


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    i always find it's best to go back to the definition:

    a set of vectors {v1,v2,...vk} is linearly independent if:

    c1v1+c2v2+...+ckvk implies:

    c1 = c2 = ... ck = 0.

    in this case, suppose that we have:

    r1a + r2b = 0.

    then we have r1a + r2b + r3c = 0

    (by setting r3 = 0), and by the ____ _____ of _______.......
  5. Feb 29, 2012 #4


    Staff: Mentor

    Had to add a couple of corrections.
    The "and there are no other solutions" part is important, because the equation c1v1+c2v2+...+ckvk = 0 has solutions whether the vectors are linearly independent or linearly dependent. The only distinction between these classifications hinges on there being a single solution (lin. independent) or an infinite number of solutions (lin. dependent), and students almost always have a hard time understanding this.
  6. Feb 29, 2012 #5


    Staff: Mentor

    No, this is absolutely not true. The vectors a, b, and c "live" in R3. It would be reasonable to say that {a, b} span a two-dimensional subspace of R3, but that is definitely not the same as R2.

    Vectors in R3 always have 3 coordinates; vectors in R2 have only two coordinates.
  7. Mar 1, 2012 #6


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    uh...i thought that that was what the word "implies" meant. i thought about using the word "forces" instead, making it clearer that the c's have to be all 0. i think this is just a semantical hitch, the idea is to set a linear combination to 0, and prove the coefficients must all be 0.

    (to say "p implies q" is the same thing as: p ONLY IF q).
  8. Mar 1, 2012 #7


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    I agree that there's no need to add that there are no other solutions, and that this follows from the meaning of "implies".
  9. Mar 1, 2012 #8


    Staff: Mentor

    Consider these vectors in R2: x1 = <1, 2> and x2 = <2, 4>.

    I set up the equation c1x1 + c2x2 = 0.

    I note that c1 = c2 = 0 satisfies the equation above, and (mistakenly) conclude that the two vectors are linearly independent.

    With P being the first equation above, and Q being the equation involving the constants, we see that P ==> Q. IOW, assuming that c1x1 + c2x2 = 0 is true, we see that c1 = c2 = 0 is true, hence the implication is true.

    This ignores the fact that there are nontrivial solutions for the constants, and this is what many students miss when they are working with the definitions of lin. independence and lin. dependence, and why I always emphasize the part that there is only one solution for the constants (the trivial solution) when the vectors are lin. independent.
  10. Mar 1, 2012 #9

    Ray Vickson

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    You are absolutely correct, but can you be sure the OP understands this? Clarification cannot hurt, and may help.

  11. Mar 1, 2012 #10


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    No, that's not true. Given that c1x1 + c2x2 = 0 we can see that c1= -2c2. c1= c2= 0 satisfy that but it does not follow that they must 0. The first equation does not imply that they are 0.

    Last edited by a moderator: Mar 1, 2012
  12. Mar 1, 2012 #11
    Mark 44 mistakenly started with an example of two linearly dependent vectors, not two linearly independent ones.
  13. Mar 1, 2012 #12
    Try these instead: For example: a=<1,0> and b=<4,3>. c1a+c2b=0 implies c1<1,0>+c2<4,3>=0 implies the system {c1+4c2=0 and 3c2=0} which clearly has no other solution besides c1=c2=0. Therefore, a and b are linearly independent.
  14. Mar 1, 2012 #13


    Staff: Mentor

    No, not mistakenly. I meant to start with two linearly dependent vectors.
  15. Mar 1, 2012 #14


    Staff: Mentor

    My whole point, and maybe I didn't state it in the best possible way is this:

    If you have an equation c1x + c2x2 + ... + cnxn = 0, then c1 = c2 = ... = cn = 0 is a solution whether or not the vectors form a linearly independent set. Over the course of many years teaching Linear Algebra, I've noticed that many students don't get this fine point, and that it helps them if you emphasize that for linearly independent vectors, the solution for the constants is unique.
  16. Mar 2, 2012 #15


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    i understand your point Mark44, and it's well-taken. as Ray pointed out, clarification doesn't hurt, as the "goal" of understanding linear independence is not to be able to state an impeccable definition of it, but rather, to be able to actually determine if sets are linearly independent or not.

    but the problem here (and i agree it IS a problem), does not lie with linear independence per se. it lies in a mistaken application of logic (which perhaps cuts deeper than the issue at hand).

    as i pointed out before, p→q means (in English) "p only if q". what happens, is people make THIS mistake:

    suppose p, instance where q is true, therefore q always true, so p→q always true.

    it is common for people to confuse inductive reasoning with deductive reasoning. the proper order of things is:

    general case implies specific case. but we often think of a specific case as being an example of the general case, and make the mistake of "over-generalizing".

    i agree that p→q and p↔q often sound "very similar". they even LOOK similar. the problem, at heart (in this definition of linear independence, and many other areas where students make "common mistakes") is not how precisely we define our terms...but rather, a failure to teach the basics of logical reasoning before throwing math (which depends on logical reasoning) proofs at students ill-equipped to understand them.
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