Are the Rows of an Invertible Matrix Linearly Independent?

[jeff1evesque]

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]

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.)