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Problem on linear independence and matrices

  1. Jan 7, 2008 #1
    Can I ask for some help?

    Suppose that {v1,v2...vn} is a linearly independent set of vectors and A is a singular matrix.
    Prove or disprove: The set {Av1, Av2, ...Avn} is linearly independent.
  2. jcsd
  3. Jan 7, 2008 #2
    How can you relate two different basis?
  4. Jan 7, 2008 #3
    the vectors are the same. and they're not bases, they're just sets of vectors
  5. Jan 7, 2008 #4
    But the vetors of the first set are linearly independent, so they form a basis. If the second set has linearly independent members it should also be a base. Correct?
  6. Jan 7, 2008 #5
    but doesnt the singular matrix change the property of linear independence?
  7. Jan 7, 2008 #6
    So the members of the second set can not be independent! :smile:
  8. Jan 7, 2008 #7
    hmm...i dont quite get it. can you write a rough proof. :) thanks a lot!
  9. Jan 7, 2008 #8
    The first set forms a basis since the vectors are linearly independent.

    Suppose now that the members of the 2nd set are linearly independent, thus they also form a basis. Two basis are related by a non-singular matrix, but in our case they related by a singular one. Thus the members of the 2nd are not linearly independent.

    What about that?
  10. Jan 7, 2008 #9
    how does the singular matrix change the linear independence of the basis?
  11. Jan 7, 2008 #10
    Let me show it with equations.

    Call the vectors of the 2nd set [itex]\bar{v}^\alpha[/itex], then

    [tex] \bar{v}^\alpha=A^\alpha_\beta\,v^\beta[/tex]

    In order for [itex]\bar{v}^\alpha[/itex] to be linearly independent, it must hold that

    [tex]\lambda_\alpha \, \bar{v}^\alpha=0\Rightarrow \lambda_{\alpha}=0 \quad \forall \alpha[/tex]

    Now we have

    [tex]\lambda_\alpha \, \bar{v}^\alpha=0\Rightarrow \lambda_\alpha \, A^\alpha_\beta\,v^\beta=0 \Rightarrow \lambda_\alpha \, A^\alpha_\beta=0 [/tex]

    The last equality hols because [itex]v^\alpha[/tex] are linearly independent.

    This is a [itex]n\times n[/itex] homogeneous system for the unknows [itex]\lambda_\alpha [/itex]. In order for that to have only the trivial solution it must hold [itex]det(A)\neq 0[/itex]. But [itex]det(A)= 0[/itex] so there is a solution for [itex]\lambda_\alpha [/itex] besides the trivial one. Thus the vectors [itex]\bar{v}^\alpha[/itex] are dependent.
    Last edited: Jan 7, 2008
  12. Jan 7, 2008 #11
    if there is a trivial solution, what does it imply? :) thanks btw.
  13. Jan 7, 2008 #12


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    How about just trying for a counter example? What is the simplest singular linear transformation you know?
  14. Jan 7, 2008 #13
    Trivial solution means [tex]\lambda^\alpha=0[/tex], which would cause [tex]\bar{v}^\alpha[/tex] to be independent
  15. Jan 17, 2008 #14

    I hope I am not spoiling the fun, but I think things are getting confused.

    Anything could happen here. Since we are not requiring that {v_1, ..., v_n} is a basis, here are two examples. Take any set of linearly independent vectors and let A = 0. Then of course their images are not linearly independent. On the other hand, take A to be some nonzero matrix, and let v_1 be any vector not in its kernel. Then {v_1} is linearly independent, and so is {Av_1}.
  16. Jan 17, 2008 #15
    The OP was:

    Since the [itex]v_i[/itex] are linearly independent, then they form a basis. Assuming of course, the dimension of the vector space is n. :smile:
  17. Jan 17, 2008 #16


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    All you really need is a 'counterexample'. If {v1, v2, ..., vn} is a set of independent vectors, and A is the linear transformation that takes every v into 0, what can you say about {Av1, Av2, ..., Avn}?
  18. Jan 17, 2008 #17


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    whats your definition of singular? if it means the columns are not independent, you are done. at least assuming you know the basic theory of dimension.
  19. Jan 18, 2008 #18


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    Yes, and masnevets' point was that there is no reason to assume that! In any case, his point was an extension of what I said: Suppose {v1, v2, ...} is a set of independent vectors and A is the zero operator (Av= 0 for all v).
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