Are there any nonhomogeneous linear systems with a solution set that forms a vector space? I know that, in order to be a vector space, a set must consists of a set V together with operations + (called (vector) addition) and · (called scalar multiplication) and an element 0 ∈ V such that all of the following properties hold:

  1. The sum of two vectors is a vector: For each v, w ∈ V , v + w ∈ V.
  2. A scalar multiple of a vector is a vector: For each c ∈ F and v ∈ V , c · v ∈ V.
  3. Vector addition is commutative: For each v, w ∈ V , v + w = w + v.
  4. Vector addition is associative: For each u, v, w ∈ V , u + (v + w) = (u + v) + w.
  5. The 0 vector is an identity for vector addition: For each v ∈ V , v+0 = 0+v = v.
  6. Every vector has an additive inverse: For each v ∈ V , there is an element w ∈ V such that v + w = w + v = 0.
  7. Multiplication by 1: For each v ∈ V , 1 · v = v.
  8. Scalar multiplication is “associative”: For each a, b ∈ F and v ∈ V , a · (b · v) = (ab) · v.
  9. Distributive law #1: For each a ∈ F and v, w ∈ V , a · (v + w) = a · v + a · w.
  10. Distributive law #2: For each a, b ∈ F and v ∈ V , (a + b) · v = a · v + b · v.

I'm uncertain what properties of the solution of a nonhomogeneous linear system would allow for connection with a vector space. I know the solution to a nonhomogeneous equation includes the solution to the associated homogeneous equation added to the particular solution, but beyond this I know very little about the properties of solution. I feel like I'm missing some key property of either vector spaces, nonhomogeneous systems or both, that is making it impossible for me to understand the question fully, let alone begin to answer it.


2 Answers 2


The solution set of Ax=b, b≠0 can never form a vector space as the zero vector ie.x=0 (additive identity) will never belong to the set of solution, and thus it fails to be a vector space.


It depends on how you look at it.

Let's work with $\mathbb{R}^n$. The solution of $Ax=b$ can be identified by $S+x_0$ for some subspace $S$ and some $x_0\in\mathbb{R}^n$. See this post.

So if you define addition on $S+x_0$ by $v+w=v+w-x_0$ and scalar multiplication by $\alpha \cdot v=\alpha(v-x_0)+x_0$, then we have a vector space. If you wish to retain the addition and multiplication from $\mathbb{R}^n$, then the $S+x_0$ is not a vector space. (It is called an affine space.)


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