If an nxn matrix has rank n how do you know it's invertible?

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Homework Help Overview

The discussion revolves around the properties of an nxn matrix with rank n and its implications for invertibility. Participants explore the relationship between rank, linear independence, and determinants within the context of linear algebra.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants consider the implications of rank on the null space and linear independence of rows. Questions are raised about the determinant and its relationship to the matrix's properties.

Discussion Status

The conversation is ongoing, with participants offering insights and questions that probe deeper into the concepts of linear independence and the dimensions of column spaces. There is no explicit consensus yet, but several productive lines of inquiry are being explored.

Contextual Notes

Participants are navigating the requirements of the homework context, including the need for proof or deeper understanding rather than just stating known results.

LeakyFrog
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Homework Statement
If an nxn matrix has rank n how do you know it's invertible?

The attempt at a solution
I know that by the Fundamental Theorem of Invertible Matrices if Rank(A) = n then A is invertible. However, I don't know if that is enough of an answer so it kind of seems like I'm supposed to prove it but I'm not sure where to start with that.

Any nudge in the right direction to get my brain cranking is appreciated.
 
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If rank(A)=n, what does that say about dim(Nul(A))?
 
If an nxn matrix has rank n, then all the rows are linearly independent from one another, so what can you say about it's determinant?
 
Let A be that matrix

Col(A) = Span(A)

dim(Col(A)) = dim(Span(A))

What dose that mean? When can the above happen, ever?
 

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