Undergrad The Paradox of u=v | Solve the Mystery

[Gjmdp]

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.

 

[fresh_42]

There is no inverse matrix for $M$ in case $n\neq m$. If $n=m$ and $M$ is regular and $Mu=Mv$ then $u=v$.

 

[Gjmdp]

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?

 

[fresh_42]

Thank you for your answer. How can you prove that?

Your reasoning is correct if those assumptions are given.

 

[Mark44]

If $n=m$ and $M$ is regular and $Mu=Mv$ then $u=v$.

I think more usual terms for regular are invertible or nonsingular.

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 $u - v = 0$