Linear dependence of square matrices

snu
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I am studying the subject of linear dependence right now and had a question on this topic. Is it possible to construct a square matrix A such that the columns of A are linearly dependent, but the columns of the transpose of A are linearly independent? My intuition tells me no, but I'm not sure how I would prove this to be the case in general. I've tried constructing several square matrices that are linearly dependent and taken the transpose and found that to be linearly dependent as well, but I'm not sure if this always holds.
 
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All matrices have the property that the determinant of the transpose is equal to the determinant of the original matrix.

From this I would gather that you can't have the property you have described above.

Look at this site:

http://mathworld.wolfram.com/Determinant.html
 
It's actually stronger than that: the column rank of any matrix (=dimension of space spanned by its columns) equals the row rank. In particular, for square matrices, if the columns are linearly dependent (column rank is smaller than matrix dimension), so are the rows.
 

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