Let #W# be a subset of a vector space #V#. If we take two vectors of #W#, then their sum is a vector of #V#, which does not necessarily lie in #W#. If we want #W# with the addition of #V# to be a vector space again, we must demand that this sum lies in #W#. A similar remark applies to the scalar multiplication of vectors from #W#.
A non-empty subset #W# of a vector space #V# is called a
linear subspace of #V# if, for all #\vec{p}#, #\vec{q}\in W# and all scalars #\lambda#, #\mu# the following holds:
\[\lambda\cdot \vec{p} + \mu \cdot \vec{q} \in W\]
- The smallest example is the subset #\{\vec{0}\}# of #V#, which only consists of the zero vector. This linear subspace is called the trivial linear subspace of #V#.
- The largest example is the subset #V# of #V#, which consists of all vectors of #V#. All other linear subspaces are called proper.
The set of real numbers, #\mathbb{R}#, with the usual addition and multiplication, is a vector space. The trivial subspace #\{\vec{0}\}# and the entire space #\mathbb{R}# are the only two linear subspaces of #\mathbb{R}#.
Each line through the origin in #\mathbb{R}^n#, that is, each subset of the form #\left.\left\{\lambda \, \vec{v}\,\right| \,\lambda\in\mathbb{R} \right\}# for a fixed non-zero vector #\vec{v}#, is a linear subspace of #\mathbb{R}^n#. If the line does not pass through the origin, it is not a linear subspace.
The requirement that #W# is non-empty, can be replaced by the requirement that the zero vector #\vec{0}# of #V# belongs to #W#. The fact is that,
- if #\vec{0}# belongs to #W#, then #W# is not empty, and
- if #W# contains a vector, say #\vec{w}#, it also contains the opposite #-\vec{w}#, and therefore also the linear combination #\vec{w}+\left(-\vec{w}\right)=\vec{0}#.
This simple property can often be used to show that a subset #W# of #V# is not a linear subspace.
This requirement that a subset is a linear subspace, guarantees that the structure of a vector space can be found on that subset.
If #W# is a linear subspace of #V#, then #W# itself, supplied with the addition and scalar multiplication of #V#, is a vector space.
The subset #L = \left.\left\{\lambda \,\rv{1,0}\,\right| \,\lambda\in\mathbb{R} \right\}# of #\mathbb{R}^2# is a linear subspace of #\mathbb{R}^2#. If we identify the point #\rv{\lambda, 0}# of #L# with the number #\lambda# of #\mathbb{R}#, then it is easy to verify that the vector space of the linear subspace #L# coincides with the vector space #\mathbb{R}#.
To see this, we must consider the eight rules for a vector space #W#. The rules dealing with equalities are satisfied because the rules already hold in #V#. This observation brings the check of rules back to those concerning the existence of the zero vector and the opposite vector of a vector #\vec{w}# in #W#.
The existence of the zero vector has already been discussed in the definition of a linear subspace.
Remains to show that, if #\vec{w}# belongs to #W#, then so does #-\vec{w}#. This follows from the fact that #-\vec{w}# is equal to the linear combination #\vec{0}+(-1)\cdot\vec{w}# of the vectors #\vec{0}# and #\vec{w}#, which both belong to #W#.
The homogeneous solution of a system of linear equations with #n# unknowns is a linear subspace of #\mathbb{R}^n#:
Consider the following general form of a homogeneous system of \(m\) linear equations with \(n\) unknowns \(x_1, \ldots, x_n\): \[\left\{\;\begin{array}{rclllllll} a_{11}x_1 \!\!\!\!&+&\!\!\!\! a_{12}x_2 \!\!\!\!&+&\!\!\!\! \cdots \!\!\!\!&+&\!\!\!\! a_{1n}x_n\!\!\!\!&=&\!\!\!\!0\\ a_{21}x_1 \!\!\!\!&+&\!\!\!\! a_{22}x_2 \!\!&+&\!\! \!\!\cdots \!\!\!\!&+&\!\! a_{2n}x_n\!\!\!\!&=&\!\!\!\!0\\ \vdots &&\vdots &&&& \vdots&&\!\!\!\!\vdots\\ a_{m1}x_1 \!\!\!\!&+&\!\!\!\! a_{m2}x_2 \!\!\!\!&+&\!\!\!\! \cdots \!\!\!\!&+&\!\!\!\! a_{mn}x_n\!\!\!\!&=&\!\!\!\!0\end{array}\right.\] Here, all \(a_{ij}\) with \(1\le i\le m, 1\le j\le n\) are real or complex numbers.
We describe a general vector in #\mathbb{R}^n# using the coordinates #x_1,\ldots,x_n#; so we view #\rv{x_1,\ldots,x_n}# as a vector of #\mathbb{R}^n#. This way, the solutions of the system equations can be viewed as a subset #S# of #\mathbb{R}^n#.
The set of solutions of the homogeneous system is a linear subspace of #\mathbb{R}^n#.
The set of solutions #\rv{x,y}# of the equation #y=0# is the line #L = \left.\left\{\lambda \,\rv{1,0}\,\right| \,\lambda\in\mathbb{R} \right\}# through the origin, i.e., a linear subspace of #\mathbb{R}^2#.
Let #W# be the set of solutions of the homogeneous system. To show that #W# is a linear subspace of #\mathbb{R}^n#, we establish the following three facts:
1. The vector #\vec{0}# belongs to #W#. If we substitute #\rv{x_1,x_2,\ldots,x_n}=\vec{0}# in the left side #a_{j,1}x_1 + a_{j,2}x_2 + \cdots + a_{j,n}x_n# of the #j#-th equation, then we find #0#, which is equal to the right hand side.
2. If #\vec{x}=\rv{x_1,x_2,\ldots,x_n}# belongs to #W# and #\lambda# is a scalar, then #\lambda\cdot \vec{x}# also belongs to #W#. This follows from \[\begin{array}{rl}&a_{j,1}\cdot\left(\lambda\cdot x_1\right) + a_{j,2}\cdot\left(\lambda\cdot x_2\right) + \cdots + a_{j,n}\cdot\left(\lambda\cdot x_n\right)\\&\phantom{xx}=\lambda\cdot\left(a_{j,1}\cdot x_1 + a_{j,2}\cdot x_2 + \cdots + a_{j,n}\cdot x_n\right)\\ &\phantom{xx}=\lambda\cdot0\\ &\phantom{xx}=0\end{array}\]
3. If #\vec{x}=\rv{x_1,x_2,\ldots,x_n}# and #\vec{y}=\rv{y_1,y_2,\ldots,y_n}# belong to #W#, then #\vec{x}+\vec{y}# belongs to #W#. This follows from \[\begin{array}{rl}&a_{j,1}\cdot\left( x_1+y_1\right) + a_{j,2}\cdot\left( x_2+y_2\right) + \cdots + a_{j,n}\cdot\left( x_n+y_n\right)\\&\phantom{xx}=\left(a_{j,1}\cdot x_1 + a_{j,2}\cdot x_2 + \cdots + a_{j,n}\cdot x_n\right)\\&\phantom{xxxx}+\left(a_{j,1}\cdot y_1 + a_{j,2}\cdot y_2 + \cdots + a_{j,n}\cdot y_n\right)\\ &\phantom{xx}=0+0\\ &\phantom{xx}=0\end{array}\]
From these three facts, it is easy to deduce that #W# satisfies the definition of a linear subspace (this conclusion is the subject of an exercise).
Later we will see that, conversely, every linear subspace of #\mathbb{R}^n# is the solution of a homogeneous system of linear equations in #n# unknowns.
Is the subset #W# of #\mathbb{R}^3# consisting of all vectors #\rv{x,y,z}# that satisfy the equation #-5\cdot x+5\cdot y+2\cdot z=-20# a linear subspace of #\mathbb{R}^3#?
No
The set #W# of solutions of the equation #-5\cdot x+5\cdot y+2\cdot z=-20# in #\mathbb{R}^3# is not a linear subspace of #\mathbb{R}^3# because the zero vector of #\mathbb{R}^3# does not satisfy the equation.
We can see this differently: the vectors #\vec{a} =\rv{4,0,0}# and #\vec{b} =\rv{0,-4,0}# belong to #W# but #\vec{a}+\vec{b}=\rv{4,-4,0}# does not, because substituting the coordinate values in the equation leads to #-40=-20#. Therefore, the linear combination #\vec{a}+\vec{b}# of #\vec{a}# and #\vec{b}# does not belong to #W#.
The notion of linear subspace
