Linear subspace/Solution space for linear system/Introduction/Section

Definition

Let $K$ be a field, and let $V$ be a $K$ -vector space. A subset $U \subseteq V$ is called a linear subspace if the following properties hold.

$0 \in U$ .
If $u, v \in U$ , then also $u + v \in U$ .
If $u \in U$ and $s \in K$ , then also $s u \in U$ holds.

Addition and scalar multiplication can be restricted to such a linear subspace. Hence, the linear subspace is itself a vector space, see exercise. The simplest linear subspaces in a vector space $V$ are the null space $0$ and the whole vector space $V$ .

Lemma

Let $K$ be a field, and let

\begin{matrix} a_{11} x_{1} + a_{12} x_{2} + \dots + a_{1 n} x_{n} & = & 0 \\ a_{21} x_{1} + a_{22} x_{2} + \dots + a_{2 n} x_{n} & = & 0 \\ ⋮ & ⋮ & ⋮ \\ a_{m 1} x_{1} + a_{m 2} x_{2} + \dots + a_{m n} x_{n} & = & 0 \end{matrix}

be a homogeneous system of linear equations over $K$ . Then the set of all solutions to the system is a linear subspace of the standard space $K^{n}$ .

Proof

See exercise.

◻

Therefore, we talk about the solution space of the linear system. In particular, the sum of two solutions of a system of linear equations is again a solution. The solution set of an inhomogeneous linear system is not a vector space. However, one can add, to a solution of an inhomogeneous system, a solution of the corresponding homogeneous system, and get a solution of the inhomogeneous system again.

Example

We take a look at the homogeneous version of example, so we consider the homogeneous linear system

\begin{matrix} 2 x & + 5 y & + 2 z & - v & = & 0 \\ 3 x & - 4 y & + u & + 2 v & = & 0 \\ 4 x & - 2 z & + 2 u & = & 0 \end{matrix}

over $ℝ$ . Due to fact, the solution set $L$ is a linear subspace of $ℝ^{5}$ . We have described it explicitly in example as

{u (- \frac{1}{3}, 0, \frac{1}{3}, 1, 0) + v (- \frac{2}{13}, \frac{5}{13}, - \frac{4}{13}, 0, 1) ∣ u, v \in ℝ} .

This description also shows that the solution set is a vector space. Moreover, with this description, it is clear that $L$ is in bijection with $ℝ^{2}$ , and this bijection respects the addition and also the scalar multiplication (the solution set $L^{'}$ of the inhomogeneous system is also in bijection with $ℝ^{2}$ , but there is no reasonable addition nor scalar multiplication on $L^{'}$ ). However, this bijection depends heavily on the chosen "basic solutions“ $(- \frac{1}{3}, 0, \frac{1}{3}, 1, 0)$ and $(- \frac{2}{13}, \frac{5}{13}, - \frac{4}{13}, 0, 1)$ , which depends on the order of elimination. There are several equally good basic solutions for $L$ .

This example shows also the following: the solution space of a linear system over $K$ is "in a natural way“, that means, independent on any choice, a linear subspace of $K^{n}$ (where $n$ is the number of variables). For this solution space, there always exists a "linear bijection“ (an "isomorphism“) to some $K^{d}$ ( $d \leq n$ ), but there is no natural choice for such a bijection. This is one of the main reasons to work with abstract vector spaces, instead of just $K^{n}$ .