View Full Version : Invertible Matrix Theorem
jeff1evesque
Aug13-09, 06:31 PM
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 A^{-1} has it's columns being linearly independent (for A^{-1}) 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
CaffeineJunky
Aug13-09, 06:53 PM
Yes. Keep in mind that \det(A) = \det(A^{T}), 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.)
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