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Rank of matrices

  1. Dec 29, 2007 #1
    The question in short is, why the rank of a matrix is equal to the rank of its transpose?

    Matrix is an array of numbers. Then it's amazing to me that the number of linear independent rows coincides with the number of linear independent columns. I tried to find some fundamental answer to this question, which does not resort to concepts like singular values or eigenvalues, so that it can be explained to those who do not know linear algebra. Is there an elementary way to explain this fact?
    Last edited: Dec 29, 2007
  2. jcsd
  3. Dec 29, 2007 #2


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    In mathematics, such "fundamental answers" are called proofs. Start searching for one. :wink:
  4. Dec 29, 2007 #3


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    No! A matrix is an object in an algebraic system with specific properties. It can be represented by an "array of numbers".

    It depends entirely upon the definition of matrix multiplication: each term in the product of two matrices is the "dot product" of a row in the first matrix with a column in the second matrix. Reversing the order enterchanges the rows and columns.
  5. Dec 29, 2007 #4
    You need several preliminary results to prove that the rank of a matrix is equal to the rank of its transpose:
    1) If A has rank r, then you can left- and right-multiply it by elementary matrices so that the rxr submatrix of the new A in the upper left corner is the identity matrix, and everywhere else is zeros.
    2) Multiplying a matrix by invertible matrices does not change its rank.
    3) Elementary matrices are invertible.
    4) The transpose of a product is the product of the transposes (in reverse order).
    5) The inverse matrix of a transpose matrix is the transpose matrix of the inverse matrix.

    Now use these results to prove the theorem.
    Last edited: Dec 29, 2007
  6. Dec 30, 2007 #5
    thank you.:smile:
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