Column, Solution and Row Spaces

  • Thread starter Thread starter jeffreylze
  • Start date Start date
  • Tags Tags
    Column Row
Click For Summary
SUMMARY

The discussion focuses on the concepts of column space, row space, and solution space for the matrix A = [1 0 1; -1 1 -3; 2 1 0]. The column space is defined by the vectors V1 = (1, -1, 2), V2 = (0, 1, 1), and V3 = (1, -3, 0), with V1 and V2 forming a basis as V3 is a linear combination of the first two. The row space consists of the vectors w1 = (1, 0, 1), w2 = (-1, 1, -3), and w3 = (2, 1, 0). The solution space is characterized as the set of column matrices X of dimension 3 by 1 such that AX = 0, and to determine if a vector is in the column or row space, one must check for a linear combination that equals the vector in question.

PREREQUISITES
  • Understanding of linear algebra concepts such as vector spaces and linear combinations.
  • Familiarity with matrix operations and augmented matrices.
  • Knowledge of basis and dimension in vector spaces.
  • Ability to perform Gaussian elimination or similar techniques to solve linear systems.
NEXT STEPS
  • Study the concept of linear independence and how to determine it for a set of vectors.
  • Learn about the Rank-Nullity Theorem and its implications for column and row spaces.
  • Explore methods for finding bases for column and row spaces using row echelon form.
  • Investigate the application of the RREF (Reduced Row Echelon Form) in solving systems of linear equations.
USEFUL FOR

Students and educators in linear algebra, mathematicians, and anyone seeking to deepen their understanding of vector spaces and their properties in the context of matrix theory.

jeffreylze
Messages
43
Reaction score
0

Homework Statement



A = [ 1 0 1
-1 1 -3
2 1 0]

what is the column space of A?

What is the row space of A

What is the solution space of A?

Also, if i were given a multiple choice, how do I check if a vector is in either the column, row or solution space of A ?

Homework Equations





The Attempt at a Solution



From my understanding, the column space is just V1 = (1,-1,2) V2 = (0,1,1) V3 = (1,-3,0) and the row space of A is w1 = (1,0,1) , w2 = (-1,1,-3) and w3 = (2,1,0) . For solution space, i have completely no idea.
 
Physics news on Phys.org
The solution space is the set of column matrices X of dimension 3 by 1 such that AX=0.

The column space is v1, v2, and v3. But are v1, v2, and v3 linearly independent? You might want to find the basis for the column space.
 
But then again, how do I check if a certain vector is in the column or row space of A?
 
The vectors form a basis. Right?
So if there exists a linear combination of the vectors that equals another 'certain vector' then the 'certain vector' is in the column or row space.

Let's see c1*v1 + c2*v2 + c3*v3 = u

Can we solve for c1, c2, and c3? If so, then u is in the space.

This looks like a job for matrices to solve. See if u=(1 1 1) is in the column space.
 
Actually only V1 and V2 forms a basis, as v3 can be written as a linear combination of V1 and V2. C1*(1,-1,2) + C2*(0,1,1) = (1,1,1) . I formed a matrix A with (1,-1,2) and (0,1,1) in columns and then put (1,1,1) into the matrix in augmented form. But the system is inconsistent, no solutions. So u is not in the column space?
 
Right on both accounts. Despite my typepo, leaving out the word 'not', I think you are seeing how all of this fits together. Well done.
 
fantastic, at least now I know I am making some progress. =D
 
hokie1 said:
The column space is v1, v2, and v3. But are v1, v2, and v3 linearly independent? You might want to find the basis for the column space.
That's not the column space. It's the subspace of R^3 (for this problem) that is spanned by v1, v2, and v3. In other words, it's the set of vectors in R^3 that are linear combinations of v1, v2, and v3.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 6 ·
Replies
6
Views
7K
Replies
8
Views
7K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
15
Views
2K