Rank of a Matrix: Can We Eliminate Lines to Get Non-Vanishing Determinant?

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If A is an nxk matrix of rank k, it is possible to eliminate n-k rows to create a new matrix A' with a non-vanishing determinant. Since the rank indicates that the row space is spanned by k vectors, the eliminated rows are linear combinations of these vectors. This reduction results in a kxk matrix A' that retains rank k, thus making it invertible. An invertible matrix has a non-zero determinant, confirming the original assertion. This concept is a key principle in linear algebra regarding matrix ranks and determinants.
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If A is an nxk matrix of real numbers (n>=k) of rank k, is it true that we can eliminate n-k lines of A to obtain a matrix A' of nonvanishing determinant?

I convinced myself of that one time while in the bus and now I can't find the proof.
 
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Hmm, if A is of rank k, then that means that the row space of A is spanned by k vectors , and this means that we can eliminate (n-k) rows of A which are effectively linear combinations of the others. So when we do that we have A', which is a kxk matrix and of rank k, which implies that it is invertible which in turn implies its det is non-zero.
 
Ohhh.. yeah!

Thanks!
 
Welcome. I'm trying to jog my linear algebra memory for a intermediate linear algebra class this semester.
 

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