Linear Algebra Matrix Proof problem

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Homework Help Overview

The problem involves proving the existence of an n x m matrix B such that the product AB equals the mxm identity matrix Im, given that A is an m x n matrix with rank m. The discussion centers around concepts in linear algebra, particularly matrix rank and properties of invertible matrices.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different approaches to the proof, including starting with definitions involving invertible matrices and considering the implications of matrix rank. Questions arise regarding notation and the interpretation of matrix products.

Discussion Status

The discussion is ongoing, with participants sharing their thoughts on various approaches and questioning the clarity of notation. Some guidance has been offered regarding the interpretation of identity and zero matrices, but no consensus has been reached on a definitive method or solution.

Contextual Notes

There are some uncertainties regarding the notation used in the problem, particularly the expression "Im 0," which has led to confusion among participants. Additionally, the assumptions about the rank of matrices and their implications are being examined.

JKCB
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Homework Statement


Let A be an m x n matrix with rank m. Prove that there exists an n x m matrix B such that AB=Im


The Attempt at a Solution


I'm assuming I would need to start with the def. That there exists P an mxm invertible matrix and Q an nxn invertible matrix s.t. A=P(Im 0)Q

then P(Im 0)Q B = Im

now I might multiply left hand side by P inverse and right hand side by Q inverse.

I'm stuck am I going in the right direction?
 
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I'm a bit confused by the notation here. What is I am 0? I'm assuming I am is the mxm identity matrix
 
Yes I am is the mxm identity matrix and 0 is the mxn zero matrix.

I'm thinking of going another direction.

What if I start with letting B be any nxm matrix with rank n. Then AB would be an mxm matrix with rank m, then by the Thm (in my book) 2.18 corollary 2 an nxn matrix is invertible iff its rank is n. AB is invertible. Let M be the inverse then (AB)M=Im which means
A(BM)=Im therefore BM would be the nxm matrix we are looking for.

What do you think? Would that do it, any holes?
 
JKCB said:
Yes I am is the mxm identity matrix and 0 is the mxn zero matrix.
As Office_Shredder asked, what does I am 0 mean? Inquiring minds want to know.
 
Im (I subscript m) is the mxm identity matrix and 0 is the m x (n-m) zero matrix.
 
Then Im 0 is the product of Im and the m x (n - m) zero matrix, which is 0. Were you thinking that Im times a zero matrix is something other than the same zero matrix?
 
No. How about this? A= (Im 0)P where P is an nxm invertible matrix. Then replace A with (Im 0)P(B) = I am then (Im 0) P P^-1(Im)=Im
(0 ) (that is a column block matrix with I am being the Identity matrix and 0 being a zero matrix (n-m) x m) then that would make
B = p^-1 (Im)
(0 ) That is a partitioned matrix I am is mxm and the zero is (n-m) x m
Will that work?
 
(that is a column block matrix with I am being the Identity matrix and 0 being a zero matrix (n-m) x m

Is this what I am 0 is? It's hard to tell
 
I think I have figured out what you're trying to communicate, but your notation was no help. What you are writing as (Im 0) looks to me like a matrix product, and what you meant was the m x n matrix (Im|0).
 

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