Column, Solution and Row Spaces

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The discussion revolves around determining the column, row, and solution spaces of the matrix A. The column space is identified as the span of vectors V1, V2, and V3, while the row space consists of the vectors w1, w2, and w3. There is a focus on checking if a specific vector is in these spaces by finding a linear combination of the basis vectors that equals the target vector. It is concluded that V1 and V2 form a basis for the column space, as V3 can be expressed as a linear combination of the first two. The conversation emphasizes understanding the relationships between these spaces and the concept of linear independence.
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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.
 
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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.
 

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