What is the proof for dim U = 3(3-r) in a 3x3 matrix over field K?

  • Thread starter Thread starter transgalactic
  • Start date Start date
  • Tags Tags
    Proof
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
Sign up Log in
3 replies · 1K views
transgalactic
Messages
1,386
Reaction score
0
there is a matrix A of M3X3 (K) over field K
so rank A=r , marked U as A of M3X3 (K) so AX=0

prove that dim U=3(3-r))
??i was told that each column gives me 3-r we have 3 columns so
it 3(3-r)

i can't understand how we get 3-r out of each column??

there is W={x exists in R^3|Ax=0}
dim W=3-r
1-1 function hase Ker (t)=0
Im (T)=WxWxW

i understand that W is the kernel of T
then they say

"dim W=3-r"
so '"r" is the Image of the matrix
but why its from one column
??
 
Last edited:
Physics news on Phys.org

Discussion

transgalactic said:
there is a matrix A of M3X3 (K) over field K
so rank A=r , marked U as A of M3X3 (K) so AX=0

prove that dim U=3(3-r))
??


i was told that each column gives me 3-r we have 3 columns so
it 3(3-r)

i can't understand how we get 3-r out of each column??
This is very hard to understand! I assume you mean that A is a 3 by 3 matrix having rank r and U is the subset of 3 by 3 matrices, X, such that AX= 0. The fact that A has rank 3 means that its image, A(K3), has dimension r. That also means that only r of its columns, thought of as vectors in K3, are independent and that kernel of A, the set of vectors, v, in K3 such that Av= 0 has dimension 3-r. If AX= 0, then X must have image a subset of the kernel of A: X must have rank less than or equal to 3-r.
 
how did they prove that
dim U=3(3-r)
??