# Subspace question? (Linear Algebra)

**Question**

Show that the solution vectors of a consistent non-homogenous system of m linear equations in n unknowns do not form a subspace of R^n

**Answer**

I know what you are talking about.

It’s in Linear Algebra.

This system can be expressed quite compactly by writing Ax=b, where A is an m x n matrix whose elements are the coefficients, aij, in the linear system. We use A to denote both the linear map and the matrix; note that A is linear map from R^n to R^m . The column vector of unknowns is x which belongs to R^n and the right-side column vector is b which belongs to R^n . When b=0, the system is said to be homogeneous.

First consider the homogeneous system Ax=0. The solution set of this system is the null space of A, denoted null A. The question is whether or not the solution set includes more than the zero vector. We have seen that if A is not injective (one-to-one), then the null space contains more than {0} . Furthermore, we have seen that if m<n, then A cannot be injective. Putting these observations together, we see that a homogeneous system with more unknowns than equations (n>m) must have nonzero solutions. This makes sense: there are not enough equations (constraints) to determine a solution uniquely.

Oops you asked about “non-homogenus solutions”. We will get there now. 🙂

Now consider the non-homogeneous system Ax=b. In order for a solution to exist, b must be a linear combination of the columns of A (with weights given by the components of x). Using familiar terminology, this means that b must lie in the range of A = R^m. Introducing new terminology that is used with matrices, this means that b must lie in the column space of A, which is the span of the columns of A. If , then there is at least one solution for all choices of b belong to R^m. This means that if A is surjective (onto), then there are solutions for all . Conversely, if n<m, then A is not surjective and no solution exists for some choices of b. We conclude that for a non-homogeneous system with more equations than unknowns (m>n), no solution exists for some choices of b. This also makes sense: it is possible that there are too many equations (constraints) to determine a solution.

What about m=n?

A special case arises when m=n. In this case, if the columns of A are linearly independent, they span R^m and all b belong to R^m lie in the column space of A (including b=0). Thus, all vectors b admit a solution, x, whose components give the weights in expressing b as a linear combination of the columns of A. Because this representation is unique, the system Ax=b has a unique solution for all b belong to R^m , in the case that m=n and the columns of A are linearly independent.

Thus,

Ax=b has solutions for ALL b. It does NOT depend on the subspace of R^n.