The Paradox of u=v | Solve the Mystery

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SUMMARY

The discussion centers on the mathematical relationship between column vectors u and v, and a matrix M. It establishes that if M is an n x n invertible matrix and M*u = M*v, then it follows that u = v. The reasoning is validated under the condition that M is regular (invertible) and n equals m. The confusion arises when considering cases where n does not equal m, as M does not possess an inverse in such scenarios.

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This discussion is beneficial for students and professionals in mathematics, particularly those studying linear algebra, as well as educators seeking to clarify concepts related to matrix theory and vector spaces.

Gjmdp
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Let u, v be column vectors n x 1 and M a m x n matrix over a field K. If M*u= M*v, then (M^-1)*M*u=(M^-1)*M*v, thus, I*u=I*v. Hence u=v. But that shouldn't be the case. What is wrong in my reasoning?
Thank you.
 
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There is no inverse matrix for ##M## in case ##n\neq m##. If ##n=m## and ##M## is regular and ##Mu=Mv## then ##u=v##.
 
fresh_42 said:
There is no inverse matrix for ##M## in case ##n\neq m##. If ##n=m## and ##M## is regular and ##Mu=Mv## then ##u=v##.
Thank you for your answer. How can you prove that?
 
Gjmdp said:
Thank you for your answer. How can you prove that?
Your reasoning is correct if those assumptions are given.
 
fresh_42 said:
If ##n=m## and ##M## is regular and ##Mu=Mv## then ##u=v##.
I think more usual terms for regular are invertible or nonsingular.
Gjmdp said:
Thank you for your answer. How can you prove that?
##Mu = Mv \Rightarrow Mu - Mv = 0 \Rightarrow M(u - v) = 0##
Assuming M is invertible, then ##M^{-1}M(u - v) = M^{-1}0 = 0##, or ##u - v = 0##
 

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