Paradox about u=v

  • I
  • Thread starter Gjmdp
  • Start date
  • #1
148
5
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.
 

Answers and Replies

  • #2
15,772
14,077
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##.
 
  • #3
148
5
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?
 
  • #5
35,431
7,295
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##
 

Related Threads on Paradox about u=v

Replies
4
Views
1K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
2
Views
381
  • Last Post
Replies
2
Views
2K
Replies
4
Views
2K
  • Last Post
Replies
10
Views
5K
  • Last Post
Replies
11
Views
4K
Replies
1
Views
3K
Replies
3
Views
3K
Top