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

Definition

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

1. ${\displaystyle {}0\in U}$.
2. If ${\displaystyle {}u,v\in U}$, then also ${\displaystyle {}u+v\in U}$.
3. If ${\displaystyle {}u\in U}$ and ${\displaystyle {}s\in K}$, then also ${\displaystyle {}su\in U}$ 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 ${\displaystyle {}V}$ are the null space ${\displaystyle {}0}$ and the whole vector space ${\displaystyle {}V}$.

Lemma

Let ${\displaystyle {}K}$ be a field, and let

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

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

Proof

${\displaystyle \Box }$

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

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

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

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

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

This example shows also the following: the solution space of a linear system over ${\displaystyle {}K}$ is "in natural way“, that means independent on any choice, a linear subspace of ${\displaystyle {}K^{n}}$ (where ${\displaystyle {}n}$ is the number of variables). For this solution space, there always exists a "linear bijection“ (an "isomorphism“) to some ${\displaystyle {}K^{d}}$ (${\displaystyle {}d\leq n}$), 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 ${\displaystyle {}K^{n}}$.