Linear Algebra 38 | Set of Solutions

For a linear system written as A x = B, the solution set is either empty or it forms an affine (shifted) subspace: once one solution exists, every other solution differs from it by a vector from the kernel of A. This matters because it turns the messy question “What are all solutions?” into two concrete tasks—find one particular solution and understand the kernel—both of which are accessible through standard linear-algebra tools.

The starting point is dimensional and geometric: A is an M×n matrix, so the linear map F_A sends vectors in R^n to vectors in R^m. The right-hand side B has M components, so hitting B is only possible if B lies in the range (image) of A. If B is not in range(A), there is no vector x that satisfies A x = B, so the solution set is empty.

When B does does lie in range(A), at least one solution exists. Pick one such solution and call it v0, meaning A v0 = B. The key observation is that the kernel of A—those vectors x0 with A x0 = 0—generates all the “ways to adjust” a solution without changing the left-hand side. Define any candidate solution as x̃ = v0 + x0. Substituting into the system gives A(v0 + x0) = A v0 + A x0 = B + 0 = B exactly when x0 is in ker(A). This yields a clean equivalence: x̃ solves A x = B if and only if x̃ can be written as v0 plus an element of ker(A).

As a result, the solution set S has the form S = v0 + ker(A) whenever solutions exist; otherwise S = ∅. In affine-subspace language, v0 + ker(A) is a translated linear subspace: it contains all vectors of the form v0 + x0 with x0 drawn from the kernel.

The discussion then connects this structure to row operations. Row operations correspond to multiplying the system on the left by an invertible matrix M, producing M A x = M B. These operations do not change the kernel of A, and they preserve the existence and form of solutions: if A v0 = B, then M A v0 = M B as well. Since both the kernel and the solvability condition remain intact, the entire solution set stays the same under row reduction.

That invariance is the bridge to Gaussian elimination. The method will be used to (1) check whether B lies in range(A) (deciding between S = ∅ and nonempty solutions), (2) produce one particular solution v0, and (3) determine a basis for ker(A). With v0 and ker(A) in hand, the full solution set S = v0 + ker(A) can be written down directly.