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Invertible Matrix Theorem

  1. Aug 13, 2009 #1
    The invertible matrix theorem states that the columns of the given matrix form a linearly independent set. Can we argue that the rows of the same matrix also forms a linearly independent set? If a matrix is invertible, then inverse of a given matrix [tex]A^{-1}[/tex] has it's columns being linearly independent (for [tex]A^{-1}[/tex]) which is equivalent to the rows of A being linearly independent. So can we say that both the rows and columns of A form a linearly independent set?

    Thanks,

    JL
     
  2. jcsd
  3. Aug 13, 2009 #2
    Yes. Keep in mind that [tex]\det(A) = \det(A^{T})[/tex], hence every portion of the invertible matrix theorem automatically applies to the rows as well as the columns. You should come to see that there is a relationship between the row space and the column space of a matrix, along with the null space, called the rank-nullity theorem.

    Note that the space spanned by the rows is different then the space spanned by the columns since row vectors live in a different vector space than the column vectors. (You will find that there is an isomorphism between the two spaces if the matrix is n x n.)
     
    Last edited: Aug 13, 2009
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