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Math Help - anyone can help me for this prove matrix

  1. #1
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    anyone can help me for this prove matrix

    1) Suppose that A is an nxn nonsingular upper triangualr matrix, and let B=A^-1. You may assume that all diagonal elements of A are nonzero.
    a) using only the definition of matrix multiplication prove that Bsubnj =0 for all j<n. ( ie, show that all elements in the nth row of B, except for the last element, are zero.)
    b) using only the definition of matrix multiplication , prove that B is upper triangular.

    Thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by happystudent View Post
    1) Suppose that A is an nxn nonsingular upper triangualr matrix, and let B=A^-1. You may assume that all diagonal elements of A are nonzero.
    a) using only the definition of matrix multiplication prove that Bsubnj =0 for all j<n. ( ie, show that all elements in the nth row of B, except for the last element, are zero.)
    b) using only the definition of matrix multiplication , prove that B is upper triangular.

    Thanks.
    Let the columns of \bold{A} be \bold{A}_1, .. \bold{A}_n, then by definition (\bold{A}_i)_i \ne 0.

    Let the k-th row of \bold{B} be \bold{B}_k.

    Then the l-th element in the k-th row of \bold{I}=\bold{B}\bold{A} is:

     <br />
\bold{I}_{k,l}=\bold{B}_k\bold{A}_l<br />

    Now, as \bold{A} is upper trianglar:

     <br />
\bold{I}_{k,1}=\bold{B}_k\bold{A}_1=(\bold{B}_k)_1 (\bold{A}_1)_1<br />

    and if k\ne 1 this is zero and as by definition (\bold{A}_1)_1 \ne 0 for k \ne 1 (\bold{B}_k)_1=0

    Now suppose that for some j<k\ (\bold{B}_k)_r=0 for all r \le j

    Then:

     <br />
\bold{I}_{k,j+1}=\sum_{r=1}^{n}(\bold{B}_k)_r (\bold{A}_{j+1})_r=\sum_{r=1}^{j+1}(\bold{B}_k)_r (\bold{A}_{j+1})_r<br />

    because \bold{A} is upper triangular, and:

     <br />
\bold{I}_{k,j+1}=\sum_{r=1}^{n}(\bold{B}_k)_r (\bold{A}_{j+1})_r=\sum_{r=1}^{j+1}(\bold{B}_k)_r (\bold{A}_{j+1})_r=(\bold{B}_k)_{j+1} (\bold{A}_{j+1})_{j+1}<br />

    because by hypothesis (\bold{B}_K)_r=0, for all r \le j.

    Hence:

     <br />
\bold{I}_{k,j+1}=(\bold{B}_k)_{j+1} (\bold{A}_{j+1})_{j+1}=0<br />

    So (\bold{B}_k)_{j+1}=0


    Which completes a proof by induction that (\bold{B}_k)_j=0 for all j<k, which proves that \bold{B} is upper triangular.

    (it is easier to see what is going on doing this element by element along the a row of \bold{B})

    RonL
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