Raising the bar

Geometry of linear equations: linear spaces and subspaces

$latex R^n$ is a linear space in the following sense: for any $latex x,y\in R^n$ and any $latex a,b\in R$ the linear combination $latex ax+by=(ax_1+by_1,...,ax_n+by_n)$ belongs to $latex R^{n}.$ From earlier exercises we know that $latex \{ ax:a\in R\}$ and $latex \{ by:b\in R\} $ are straight lines drawn through the vectors $latex x,y.$ Therefore $latex \{ ax+by:a,b\in R\}$ is a plane drawn through those vectors. We can say that $latex R^n$ is a linear space in the following sense: for any $latex x,y\in R^n$, the whole plane drawn through $latex x,y$ is contained in $latex R^n.$

Subsets of $latex R^n$ may have this property.

Example 1. On the plane, take any straight line passing through the origin (the slope doesn't matter). The equation of such a line is

(1) $latex ax+by=0,$

where at least one of the coefficients $latex a,b$ is different from zero. If $latex b\neq 0,$ we can solve the equation for $latex y$ and get a more familiar form $latex y=kx.$ If $latex b=0,$ we get a vertical line $latex x=0.$ If two points $latex (x_1,y_1),$ $latex (x_2,y_2)$ satisfy (1), then you can check that their linear combination $latex c(x_1,y_1)+d(x_2,y_2)$ also satisfies (1). A straight line that does not pass through the origin is described by $latex ax+by=c$ with $latex c\neq 0.$

Example 2. In $latex R^3$, take any straight line or any 2D plane passing through the origin. A straight line is described by a system of two equations $latex a_1x+b_1y+c_1z=0,$ $latex a_2x+b_2y+c_2z=0$ and a plane is described by one equation $latex ax+by+cz=0.$ You can do the algebra as above to show that linear combinations of elements of these sets belong to these sets. However, I suggest using the geometric interpretation of linear operations to show that these are examples of subspaces.

Figure 1. A hyperplane is not a subspace

A straight line or a plane that does not pass through the origin is not a subspace, see Figure 1. The straight line $latex L$ does not pass through the origin. If we take vectors $latex X,Y$ from it, their sum, found by the parallelogram rule, does not belong to $latex L$.

Definition 1. The set $latex L\subset R^n$ is called a linear subspace of $latex R^{n}$ if for any $latex x,y\in L$ it contains the whole plane drawn through $latex x,y.$ By induction, any linear combination $latex a_1x^{(1)}+...+a_mx^{(m)}$ of any elements of a subspace belongs to it.

The definition does not exclude the extreme cases $latex L=\{0\}$ and $latex L=R^n$. These two subspaces are called trivial.

Sometimes we are interested in how a subspace is generated.

Definition 2. Take any vectors $latex x^{(1)},...,x^{(k)}\in R^n$ and consider the set $latex L$ of all linear combinations

(2) $latex a_1x^{(1)}+...+a_kx^{(k)}$ with $latex a_1,...,a_k\in R.$

This set is a linear subspace (because a linear combination of expressions of type (2) is again of this type) and it is called a span of $latex x^{(1)},...,x^{(k)}.$ We also say that $latex L$ is spanned by $latex x^{(1)},...,x^{(k)}$ and use the notation $latex L=span[x^{(1)},...,x^{(k)}].$

In this terminology, the first characterization of the matrix image says that it is spanned by that matrix' columns.

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