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

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Discussion Overview

The discussion revolves around the properties of a matrix A of rank k, specifically whether it is possible to eliminate n-k rows to obtain a new matrix A' with a non-vanishing determinant. The focus is on linear algebra concepts related to matrix rank and determinants.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant questions whether it is true that eliminating n-k rows from a rank k matrix A can yield a matrix A' with a non-vanishing determinant.
  • Another participant explains that since A has rank k, its row space is spanned by k vectors, allowing for the elimination of (n-k) rows that are linear combinations of the others, leading to a kxk matrix A' that is invertible and thus has a non-zero determinant.
  • A later reply expresses agreement with the explanation provided.
  • Another participant mentions their intention to refresh their memory on linear algebra concepts for an upcoming class.

Areas of Agreement / Disagreement

Participants appear to agree on the reasoning that eliminating certain rows from a rank k matrix can lead to a matrix with a non-vanishing determinant, though the initial question remains open for further exploration.

Contextual Notes

The discussion does not delve into specific proofs or detailed conditions under which the elimination of rows leads to a non-vanishing determinant, leaving some assumptions and mathematical steps unresolved.

Who May Find This Useful

Students and individuals interested in linear algebra, particularly those studying matrix theory and determinants.

quasar987
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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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