Does the set of solutions for Ax=b form a subspace?

In summary: So xparticular cannot be a linear combination of any vectors in the nullspace.In summary, the solution space of A x = b does not form a subspace.
  • #1
mitch_1211
99
1
I have a non-homogeneous Ax=b (with b non-zero) and i want to know if the set of all the solution vectors, x, forms a subspace.

I know that every solution can be written as x = xparticular + xhomogeneous i.e as the sum of a particular solution and a homogeneous solution, but I'm not sure if this helps here...

thanks!
 
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  • #2
Just try it:

Let x be a solution and c a constant, then
A (c x) = c A x = c b != b

Let x1 and x2 be solutions, then
A (x1+x2) = A x1 + A x2 = b + b = 2b != 0

So no, the solutions do not define a linear subspace.

---------------------------------------------------------------

That said, they do define a nice plane within the original vector space.
This plane is actually an affine space (http://en.wikipedia.org/wiki/Affine_space)
In the words of the French mathematician Marcel Berger,
"An affine space is nothing more than a vector space whose origin we try to forget about, by adding translations to the linear maps"

By choosing a fixed vector in the affine space you can recover the linear structure.
Let x0 be any fixed vector in the solution space of A x0 = b.
Then define for any x st A x = b, define y = x - x0, which satisfies
A y = A x - A x0 = b - b = 0.
The set of all { y = x - x0 | A x = b } forms a nice linear space.
The y are in 1:1 correspondence with the x,
so we can redefine addition and multiplication on the original x and recover a linear space:
c1 * x1 + c2 * x2 = c1 x1 + c2 x2 - (c1 + c2 - 1) x0
The solution space of A x = b is linear with respect to the new, underlined multiplication and addition.
Note that x0 is the zero vector wrt the underlined operations
x + c * x0 = x + c x0 - (1 + c - 1) x0 = x
This is basically what's said in http://en.wikipedia.org/wiki/Affine_space#Affine_subspaces
 
  • #3
Besides the checking above, there is an explanation why it is not so.

From x = xparticular + xhomogeneous,

3 things i want to mention:
1)xparticular is in the row space of A.
2)xhomogeneous is in the nullspace of A.
3)xparticular is unique.

so xparticular is just 1 vector(cannot form a subspace) and this vector is independent of any vector in the nullspace of A. This is because row space of A and the nullspace of A is orthogonal.
 

1. What is a subspace?

A subspace is a subset of a vector space that is closed under addition and scalar multiplication. This means that any linear combination of vectors within the subspace will also be within the subspace.

2. How can I determine if Ax=b forms a subspace?

To determine if Ax=b forms a subspace, we need to check if it satisfies the two conditions of a subspace: closure under addition and scalar multiplication. This means that if we add two solutions together or multiply a solution by a scalar, the resulting vector must also be a solution to the equation.

3. What is the importance of determining if Ax=b forms a subspace?

Determining if Ax=b forms a subspace is important because it tells us if the set of solutions to the equation forms a vector space. This can help us in solving other problems and understanding the properties of the solutions.

4. Can Ax=b form a subspace if the matrix A is not invertible?

No, if the matrix A is not invertible, then the set of solutions for Ax=b will not form a subspace. This is because if A is not invertible, there will be infinitely many solutions to the equation, making it impossible for the set to be closed under addition and scalar multiplication.

5. How does the dimension of the matrix A affect the subspace formed by Ax=b?

If the matrix A is a square matrix with full rank, then the dimension of the subspace formed by Ax=b will be equal to the number of columns in A. However, if A is not full rank, then the dimension of the subspace will be less than the number of columns in A.

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