Solution x spanning columns of A

  • Thread starter Thread starter fackert
  • Start date Start date
  • Tags Tags
    Columns
fackert
Messages
3
Reaction score
0
This is a really odd question to me. Usually we talk about b (in Ax=b) spanning the columns of A but here its talking about x spanning A.

If x=<-3,4,8> (this is a vertical 3x1 matrix) is a solution to Ax=0, then is x in the set spanned by the columns of A?

I'm pretty sure it is no, but i can't explain why? (briefly). And i don't know how to change the statement so it would be true.
 
Physics news on Phys.org
One obvious point- if A is an m by n matrix (m rows, n columns), then A maps an Rn to Rm. In order that we be able to multiply A times x, x must have n rows and so must be in Rn. But the columns each have m elements, one for each of the n rows, and so span a subspace of Rm. Members of the "column space" are vectors in the image of A, vectors of the form y= Ax for some x, not x itself.
 
Actually if its a solution to the matrix equation Ax=0 that's considered a nullspace.
 
Thanks, Halo31. fackert, if A maps vector space U to a subspace of vector Space V, then all solutions of Ax= 0, the "null space" of A, are in U. The columns of A, thought of as vectors (the "column space" of A), are in V. Of course, any x that A can be applied to must be in U, not V.
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
Back
Top