Results 1 to 11 of 11

Math Help - Prove A is singular

  1. #1
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128

    Prove A is singular

    Let A be an nxn marix, if A is row equivalent to a matrix B and there is a non-zero matrix C such that BC = 0n, prove that A is singular.
    Last edited by 450081592; November 15th 2009 at 01:55 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by 450081592 View Post
    Let A be an nxn marix, if A is row equivalent to a matrix B and there is a non-zero matrix C such that BC = 0n, prove that A is singular.

    A row equivalent to B \Longrightarrow\,\exists elementary matrices E_1,...,E_k\,\,s.t.\,\,E_1...E_kA=B , so:

    0_n=BC=(E_1....E_kA)C , and if A weren't singular and since elementary matrices aren't singular, multiplying the right hand from the left by A^{-1}E_K^{-1}...E_1^{-1} you get a contradiction.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    Quote Originally Posted by tonio View Post
    A row equivalent to B \Longrightarrow\,\exists elementary matrices E_1,...,E_k\,\,s.t.\,\,E_1...E_kA=B , so:

    0_n=BC=(E_1....E_kA)C , and if A weren't singular and since elementary matrices aren't singular, multiplying the right hand from the left by A^{-1}E_K^{-1}...E_1^{-1} you get a contradiction.

    Tonio

    if A weren't singular and since elementary matrices aren't singular, multiplying the right hand from the left by A^{-1}E_K^{-1}...E_1^{-1} you get a contradiction.

    what does does this mean? are we not trying to prove that A is singular? there shouldnt be a contradction?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by 450081592 View Post
    if A weren't singular and since elementary matrices aren't singular, multiplying the right hand from the left by A^{-1}E_K^{-1}...E_1^{-1} you get a contradiction.

    what does does this mean? are we not trying to prove that A is singular? there shouldnt be a contradction?

    Read again and carefully: we supposed A is NON-SINGULAR, i.e. we supposed A is invertible, and we got a contradiction...and this is what you wanted, didn't you?

    Tonio
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    Quote Originally Posted by tonio View Post
    Read again and carefully: we supposed A is NON-SINGULAR, i.e. we supposed A is invertible, and we got a contradiction...and this is what you wanted, didn't you?

    Tonio
    oh ic, so we are trying to prove that A is not an elementaery matrix, right?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    Quote Originally Posted by tonio View Post
    A row equivalent to B \Longrightarrow\,\exists elementary matrices E_1,...,E_k\,\,s.t.\,\,E_1...E_kA=B , so:

    0_n=BC=(E_1....E_kA)C , and if A weren't singular and since elementary matrices aren't singular, multiplying the right hand from the left by A^{-1}E_K^{-1}...E_1^{-1} you get a contradiction.

    Tonio

    E_1,...,E_k\,\,s.t.\,\,E_1...E_kA=B , this line, isn't it Ek...E2E1A = B? does it make a different? And what does multiplying the right hand from the left mean, multiply that at the right side, or left side?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by 450081592 View Post
    oh ic, so we are trying to prove that A is not an elementaery matrix, right?

    No. We're trying to prove that A is singular. Of course, this includes A not being elementary since elem. matrices are non-singular, but our quest is way more general.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by 450081592 View Post
    E_1,...,E_k\,\,s.t.\,\,E_1...E_kA=B , this line, isn't it Ek...E2E1A = B? does it make a different? And what does multiplying the right hand from the left mean, multiply that at the right side, or left side?

    I decided E_1...E_kA. but of course you can order the elem. matrices as you wish, and yes: multiplying from the left means at the left side, since matrix multiplication isn't commutative.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    Quote Originally Posted by tonio View Post
    A row equivalent to B \Longrightarrow\,\exists elementary matrices E_1,...,E_k\,\,s.t.\,\,E_1...E_kA=B , so:

    0_n=BC=(E_1....E_kA)C , and if A weren't singular and since elementary matrices aren't singular, multiplying the right hand from the left by A^{-1}E_K^{-1}...E_1^{-1} you get a contradiction.

    Tonio

    ok, but why do I multiply [tex]A^{-1}E_K^{-1}...E_1^{-1}[/math? where is it from, is it the inverse of B? If I only multiply this on the left side, won't it change the equation?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by 450081592 View Post
    ok, but why do I multiply A^{-1}E_K^{-1}...E_1^{-1}? where is it from, is it the inverse of B? If I only multiply this on the left side, won't it change the equation?

    Do it! Multiply the expression BC=(E_1...E_kA)C=0 from the left by A^{-1}E_k^{-1}...E_1^{-1} and you'll get a contradiction....don't ever forget the given data!

    Tonio
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    ok I got it, we multiply by the inverse of B, that we get C =0 which contradicts the hypothesis, thanks so much
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. singular matrices prove
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: June 26th 2011, 01:14 AM
  2. Prove that matrix A is singular
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: February 7th 2011, 09:38 PM
  3. singular
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: November 15th 2010, 07:04 AM
  4. Prove that a matrix is non-singular
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: February 16th 2009, 02:06 PM
  5. Replies: 3
    Last Post: September 5th 2007, 07:43 AM

Search Tags


/mathhelpforum @mathhelpforum