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urgent! matrix help!

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    • Thread Starter

    let A =

    i need to find a 5x5 matrix M with rank(M) = 2 that AM = 4X5 zero matrix.
    I thought of finding the inverse of A at first, but A is not a square matrix!!! i got stuck... help pls thanks!

    and for any 5X5 matrix B such that AB = 0 then rank B <= 2. how can i prove this?
    • 1 follower

    For the first bit, since rank(M) = 2 why don't you try:

    M = \begin{pmatrix} a & b & c & d & e \\f & g & h & i & j \\0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 \end{pmatrix}?

    Then you could sub it in AM = 0 to find relations on a-j and just choose any old matrix that works.
    • Thread Starter

    omg are you saying i have to multiply that all out and solve for it? is there a proper way of doing it?

    That is the proper way of doing it. Pick any two linearly independent solutions of the system Ax=0. Then let B be the matrix whose first two columns are the solutions, and fill the remaining columns with zeros.

    You can do the second bit easily if you know the following inequality. If A is an mxn matrix, and B is an nxp matrix, then
    rank(A) + rank(B) - n <= rank(AB)

    (Try to justify it by drawing a diagram or something, if you don't feel like proving it carefully.)

    So in this case,
    3 + rank(B) - 5 <= 0
    => rank(B) <= 2


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Updated: March 28, 2007
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