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Singular / nonsingular question

  1. Sep 22, 2004 #1
    Please help me on this....!!!

    Let A and B be n x n matrices and let C = AB. Prove that if B is singular then C must be singular.
    (The hint they provided is to use this theorem : Equivalent conditions for nonsingularity. 1. A is singular. 2. A x=0 has only the trivial soultion 0. 3. A is row equivalent to I )

    Please please help!!!
     
  2. jcsd
  3. Sep 22, 2004 #2

    matt grime

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    So, if B is singular there is a nonzero x such that Bx=0, and, since for all matrices M0=0, then what is Cx?
     
  4. Sep 23, 2004 #3
    Oh, I understand now. Thanks~
     
  5. Sep 23, 2004 #4

    mathwonk

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    did you try that problem befiore asking for help?
     
  6. Sep 24, 2004 #5
    of coz i did, but did it in a wrong direction by dealing with the inverses.
     
  7. Sep 26, 2004 #6

    matt grime

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    well, that also works: if C has an inverse then TC=1, say, so that TAB=1, hence B ins invertible, with inverse TA, contradiction, so C cannot posses an inverse.
     
  8. Sep 26, 2004 #7

    mathwonk

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    here is why i asked what i did.

    you wrote:
    C = AB. Prove that if B is singular then C must be singular.
    (The hint they provided is to use this theorem : Equivalent conditions for nonsingularity. 1. A is singular. 2. A x=0 has only the trivial solution 0."


    AHA!! now I see your problem, you have the definitions backwards! this is the definition of non singular, rather than the definition of singualr. you could not possibly do the problem with this incorrect version of the notion of singular. i.e. 1 is not what it should be. was that the rpoblem?

    if notm and you really understood that non singular meant that Ax = 0 impleis x =0, then,

    1) did you understand that then B singualr means that there is some x which is not zero but with Bx = 0?

    If so, then it is almost trivial to see that also (AB)x= A(Bx) = A0 = 0 , hence AB is singular.

    I am having difficulty thinking you did not see how go from Bx = 0 to ABx = 0, so I was thinking the problem was elsewhere. Maybe in stating the definition of singular?

    If I am wrong, then notice that if you see that Bx = 0,ma nd you are asking yourself if (AB)y everye quals zero for a non zero y, that x is a natural choice. Indeed it is the only choice you have from what is given. As in zenk, you must use whatever you are given.

    The reason I ask is it is always helpful to see where you went astray, so as to observe how to avoid it next time.
     
  9. Mar 30, 2009 #8
    Do you mean with TC
    [tex] C^T,[/tex]
    with TAB
    [tex] A^{T}B[/tex]
    and with TA
    [tex] A^T[/tex]?
     
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