Linear Algebra: Proving AB Not Invertible for mXn Matrix

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Homework Help Overview

The discussion revolves around the properties of matrix multiplication, specifically focusing on the conditions under which the product of two matrices, AB, is not invertible. The matrices in question are defined as A being an mXn matrix and B an nXm matrix, with the condition that n is less than m.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the dimensions of matrices A and B on the invertibility of their product AB. There are attempts to demonstrate that AB is not linearly independent and to identify nonzero vectors that could lead to AB not being invertible. Questions arise regarding the presence of free variables in matrix B and how that affects the overall invertibility of AB.

Discussion Status

The discussion is ongoing, with participants offering insights into the implications of linear dependence and the existence of nonzero solutions to equations involving AB. There is a focus on understanding the relationship between the dimensions of the matrices and the conditions for invertibility, though no consensus has been reached yet.

Contextual Notes

Participants are navigating the constraints imposed by the dimensions of the matrices, particularly the fact that n is less than m, which raises questions about the linear independence of the resulting product matrix AB.

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Homework Statement


If A is an mXn matrix, B is an nXm matirx, and n<m, then AB is not invertible.


Homework Equations





The Attempt at a Solution


By doing A is a 2X1 and B is a 1X2, I find that AB is not linearly independent, so it cannot be invertible, but I'm not sure how to show that for all matrices of this nature.
 
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Can you show there is a nonzero vector x such that Bx=0? That would make big problems for AB being invertible. And don't PM people about problems, ok? Just post it on the forums and wait a bit.
 
Since n<m, there will be a free variable in the nXm matrix B when reduced to echelon form, correct? So then there is obviously more than the trivial solution.
I'm still confused as to why that creates a problem for AB being invertible.
 
If AB has an inverse (AB)^(-1), then for every x, (AB)^(-1)*ABx=x. What happens if ABx=0 and x is not zero?
 

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