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Math Help - matrix and determinants

  1. #1
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    matrix and determinants

    Let A be an m x n matrix, where m<n, then A^t(A) is an n by n matrix. show that det(A^t(A))=0

    also if A is a 2 x 3 matrix can you say that det (AA^t)=0

    thanks
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  2. #2
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    Quote Originally Posted by b0mb3rz View Post
    Let A be an m x n matrix, where m<n, then A^t(A) is an n by n matrix. show that det(A^t(A))=0

    also if A is a 2 x 3 matrix can you say that det (AA^t)=0

    thanks
    I'm not the best at Linear Algebra but I'll give this a crack until someone else helps you out.

    I'm assuming A^t(A) to mean the transpose matrix of A. So if A is an m x n matrix then the transpose should be n x m, since the rows and columns switch.

    Before I try to help, is this what you meant?
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  3. #3
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    yeah it is and it says hint, consider the system of homogeneous linear equations Ax=0
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  4. #4
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    Quote Originally Posted by b0mb3rz View Post
    yeah it is and it says hint, consider the system of homogeneous linear equations Ax=0
    Well determinants only apply to square matrices, so I don't see how this problem is valid.
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  5. #5
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    Quote Originally Posted by Jameson View Post

    Well determinants only apply to square matrices, so I don't see how this problem is valid.
    b0mb3rz didn't probably make it clear but i think what he meant was \det (A^TA)=0. this is not difficult to prove: \text{rank}(A^TA) \leq \min \ (\text{rank}A^T, \text{rank}A) \leq \min \ (m,n)=m < n. \ \ \ \color{red}(*)

    thus, by the rank-nullity theorem, the null-space of A^TA is non-zero and hence \det (A^TA) = 0.


    \color{red}(*) you may also argue that: \text{rank}(A^TA)=\text{rank}A \leq \min(m,n)=m < n.
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