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

Definition

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

1. ${}0\in U$.
2. If ${}u,v\in U$, then also ${}u+v\in U$.
3. If ${}u\in U$ and ${}s\in K$, then also ${}su\in U$ 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

${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&0\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&0\\\vdots &\vdots &\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}&=&0\end{matrix}}$

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.

$\Box$

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 to the inhomogeneous system again.

Example

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

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

over ${}\mathbb {R}$. Due to

the solution set ${}L$ is a linear subspace of ${}\mathbb {R} ^{5}$. We have described it explicitly in example as

${\left\{u{\left(-{\frac {1}{3}},0,{\frac {1}{3}},1,0\right)}+v{\left(-{\frac {2}{13}},{\frac {5}{13}},-{\frac {4}{13}},0,1\right)}\mid u,v\in \mathbb {R} \right\}},$

which also shows that the solution set is a vector space. With this description, it is clear that ${}L$ is in bijection with ${}\mathbb {R} ^{2}$, and this bijection respects the addition and also the scalar multiplication (the solution set ${}L'$ of the inhomogeneous system is also in bijection with ${}\mathbb {R} ^{2}$, but there is no reasonable addition nor scalar multiplication on ${}L'$). However, this bijection depends heavily on the chosen "basic solutions“ ${}{\left(-{\frac {1}{3}},0,{\frac {1}{3}},1,0\right)}$ and ${}{\left(-{\frac {2}{13}},{\frac {5}{13}},-{\frac {4}{13}},0,1\right)}$, 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 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 for is no natural choice for such a bijection. This is one of the main reasons to work with abstract vector spaces, instead of just ${}K^{n}$.