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

[mitch_1211]

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 = x_particular + x_homogeneous i.e as the sum of a particular solution and a homogeneous solution, but I'm not sure if this helps here...

thanks!

 

[Simon_Tyler]

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.

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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

 

[td21]

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.