The notion of linear subspace

Vector spaces: Vector spaces and linear subspaces

The notion of linear subspace

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 $W$ .

Linear subspace
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.

Examples

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.

Linear subspaces are vector spaces

If $W$ is a linear subspace of $V$ , then $W$ itself, supplied with the addition and scalar multiplication of $V$ , is a vector space.

Example 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 $\mathbb{R}^n$ .

Example 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 $-x+3\cdot y+3\cdot z=6$ a linear subspace of $\mathbb{R}^3$ ?

No

The set $W$ of solutions of the equation $-x+3\cdot y+3\cdot z=6$ 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 $\vec{a} =\rv{-6,0,0}$ and $\vec{b} =\rv{0,2,0}$ belong to $W$ but $\vec{a}+\vec{b}=\rv{-6,2,0}$ does not, because substituting the coordinate values in the equation leads to $12=6$ . Therefore, the linear combination $\vec{a}+\vec{b}$ of $\vec{a}$ and $\vec{b}$ does not belong to $W$ .

New example