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Let A be an mxn matrix with columns C1, C2,...,Cn...?
If rank A=n, show that {A^TC1, A^TC2,...,A^TCn} is a basis of R^n
1 Answer
- nyc_kidLv 79 years agoFavorite Answer
Here is a useful, well known fact (I don't prove well known facts) : If you multiply A by its transpose, then, the resulting square matrix has the same rank as A.
Now, {A^TC1, A^TC2,...,A^TCn} are just the columns of the matrix B = (A^t)A
And since rank A = n , so is rank B. So the n columns of B ({A^TC1, A^TC2,...,A^TCn}) are linearly independent , and any n independent vectors in R^n constitute a base.