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Math Help - Square of a matrix...

  1. #1
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    Square of a matrix...

    Hey, I did this last semister in Matrix Algebra, and now I need it for a proof in Lineary, but can't remember how it goes (and annoyingly can't find my notes).

    If A is an n by n matrix, and A^2 = 0, does this force A = 0? My gut tells me no...but I can't come up with a counter example.


    This isn't the actual question I'm facing, but I think it might help in my serach for the answer. The question I'm doing says "Show A^2 = 0 iff col A is contained within null A.

    I'm not sure if I'm on the right track with the whole "Is A forced to be 0 thing", but it was the best I could think of to start it off.

    Thanks!
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  2. #2
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    I worked it out but not in full detail and I think the answer is that A=0.

    For example say,
    [a b]
    [c d] =A
    Then solve,
    [0 0]
    [0 0] = A^2
    Meaning,
    a^2+bc=0
    ad+bd=0
    ac+cd=0
    bc+d^2=0
    I think the only solution is the trivial one.
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  3. #3
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    Hello, Richard!

    If A is an n-by-n matrix, and AČ = 0, does this force A = 0? . no
    [size=3]
    Code:
    We have:   | a b | | a b |     | 0 0 |
               |     | |     |  =  |     |
               | c d | | c d |     | 0 0 |
    
    
    Then:  [1] aČ + bc = 0    [2] ab + bd = 0
           [3] ac + cd = 0    [4] bc + dČ = 0

    From [2], we have: .b(a + d) .= .0
    From [3], we have: .c(a + d) .= .0

    If b,c ≠ 0, then: .d = -a


    From [1], we have: . .= .-bc

    Since aČ is positive, b and c must have opposite signs.
    And a is the mean proportional of |b| and |c|.

    Some examples:
    Code:
        | 4 -2 |       | 3  0 -1 |
        | 8 -4 |       | 0  0  0 |
                       | 9  0 -3 |
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  4. #4
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    Thanks a million! This has been bugging me all day. Guess I'm back to square one with my original problem though ....Though I did notice that the rows and columns of those two matrices are linearly dependant....hmm, that might be useful....
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  5. #5
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    Quote Originally Posted by Richard Rahl View Post
    This isn't the actual question I'm facing, but I think it might help in my serach for the answer. The question I'm doing says "Show A^2 = 0 iff col A is contained within null A.
    Let A be a square nxn matrixm the null(A) is the space of col.
    vectors x, such that Ax=0.

    col(A) is the space spanned by the cols of A.

    Let A_i be the i'th col of A, then:

    A^2=A[A_1,A_2, .., A_n]=0

    so for each i, A A_i =0, so A_i is in the null space of A. Hence col(A) is
    a subspace of null(A)

    Similarly if col(A) is a subspece of null(A), then A A_i=0 for each col A_i of A,
    but:

    A^2=AA=A[A_1,A_2... A_n]=0.

    Which completes the proof.

    RonL
    Last edited by CaptainBlack; February 13th 2007 at 09:53 AM.
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  6. #6
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    Okay, just to recap and make sure I understand:

    A^2 = A[C1....Cn] = 0

    In order for the two matricies to equate (A^2 = the zero matrix), when we multiply each column of A by A, we get:

    [AC1 AC2 .... ACn] (distributive)

    since this equals the zero matrix, each ACi = 0

    So, since null A is the set of solution vectors to AX=0, which by default always has the trivial solutionh, given that every element of the resulting matrix is 0, it's colums must be contained within null A (its always a subset, since they are all zeros, which is always in null A).

    Given that it's an IFF statement, just turn this around...

    col A c null A -> A^2 = 0

    Given that col A is a subset of null A:

    each Ci is an element of null A

    if so, each column of ACi = 0

    which is the same as

    0 = A[C1....Cn]

    which is

    0 = AA

    0 = A^2

    QED


    (PS...can a non trivial solution exist such that such that col B c null A?, I don't think so???)
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by Richard Rahl View Post
    Okay, just to recap and make sure I understand:

    A^2 = A[C1....Cn] = 0

    In order for the two matricies to equate (A^2 = the zero matrix), when we multiply each column of A by A, we get:

    [AC1 AC2 .... ACn] (distributive)

    since this equals the zero matrix, each ACi = 0

    So, since null A is the set of solution vectors to AX=0, which by default always has the trivial solutionh, given that every element of the resulting matrix is 0, it's colums must be contained within null A (its always a subset, since they are all zeros, which is always in null A).

    Given that it's an IFF statement, just turn this around...

    col A c null A -> A^2 = 0

    Given that col A is a subset of null A:

    each Ci is an element of null A

    if so, each column of ACi = 0

    which is the same as

    0 = A[C1....Cn]

    which is

    0 = AA

    0 = A^2

    QED

    A bit long winded but OK

    RonL
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