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Linear Algebra Proof

  1. Mar 1, 2007 #1


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    1. The problem statement, all variables and given/known data
    Prove the following theorm:
    Given any nxn matrix A, a mxm matrix B, the m+n x m+n matrix C:

    where the top left corner of C is A, the top right (X) is any mxn matrix, the bottom right is a nxm 0 matrix and the bottom right is B.
    |C| = |A|*|B| (|C| is the determinant of C...)

    I found a proof but the book did something different and i want to check if my way is correct.

    2. Relevant equations
    1) any square matrix that can't be reduced to the I matrix has a determinant of 0. (because it can be reduced to a matrix with a 0 row or coloum)
    2) multiplying a row or coloum by a scalar changes the determinant by a factor of 1 over the scalar.
    2) changing rows changes the sign of the determinant
    3) adding the multiple of a row doesn't change the D.

    3. The attempt at a solution

    Here's my proof:
    Lets say that both A and B can be reduced to the I matrix. if A can be reduced than it can be done with by operating only with the coloums using the 3 basic operations. the process of the reduction will change the D of A (|A|) by some factor that we'll call 'a'. so the determinant of A is just a (|A| = a) - this is because the determinant of I is 1.
    The same argument can be made with B - this time operating on the rows giving the determinant of b.
    Since we reduced A by operating on the coloums and B with the row, we can do all those operations on C without the operation on one of them effecting the rest of the matrix. Applying these operations to C changes the determinant by a*b and since the resaulting matrix is a trianguler matrix with a diagonal of only ones - the determinant of C is a*b. And so:
    |C| = a*b = |A| * |B|.

    If either A or B can't be reduced then one of them has a 0 determinant so |A|*|B| = 0
    and by an argument similar to the above |C| can also be reduced to a matrix with either a 0 row or 0 coloum so |C| = 0.
    Is this a correct proof?
  2. jcsd
  3. Mar 1, 2007 #2


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    Really? And what if A = I ? :smile:

    Try to use induction, I found the proof more elegant, but then again, degustibus.
  4. Mar 1, 2007 #3


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    Yeah, the proof looks correct. You could clean it up a little though. There is an invertible matrix U such that AU is either identity (if A is reducible) or has a zero column (if A is not). There is also an invertible V such that VB is either identity or has a zero row. The matrix


    which we'll call U', clearly has determinant |U|, and


    which we'll call V', has determinant |V|. If A has a zero column, then V'CU' does too. If B has a zero row, then V'CU' does too. So if either of A or B has determinant 0, V'CU' does too, so |V'CU'| = 0. Since V' and U' are invertible, this means |C| = 0, as desired. If on the other hand both A and B have non-zero determinant, then it's clear that V'CU' is just


    which has determinant 1. So

    |V'CU'| = 1
    |V'||C||U'| = 1
    |C| = (|V'||U'|)-1
    |C| = (|V||U|)-1

    Since AU = I = VB, U and V are just the inverses of A and B, thus have reciprocal determinants to A and B, i.e. |U| = |A|-1, |V| = |B|-1. This gives the desired results.

    Your argument was a really good one, I was just thinking it's better to clean it up instead of saying things like

    "the process of the reduction will change the D of A (|A|) by some factor that we'll call 'a'."

    which are unclear, and in fact wrong in this case (the determinant of A never changes, but row reducing A creates a new matrix with a new determinant).
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