Section 2.2 Linear Spans
In this section we define linear span of a subset of \(\R^n\) and look at several example. We also look at an important class of suspaces associated with a matrix.
Definition 2.2.1.
Let \(S=\{v_1, v_2,\ldots, v_k\}\) be a non empty subset of vectors in \(\R^n\text{.}\) Then the linear span of \(S\text{,}\) denoted by \(L(S)\) is a subset of \(\R^n\) defined as
\begin{equation*}
L(S):=\{\alpha_1v_1+\alpha_2v_2+\cdots+\alpha_kv_k:\alpha_1,\cdots,\alpha_k\in \R\}.
\end{equation*}
Note that if \(x\in L(S)\) means there exist scalars \(\alpha_1,\ldots,\alpha_k \in \R\) such that \(x=\alpha_1v_1+\alpha_2v_2+\cdots+\alpha_kv_k\text{.}\)
Example 2.2.2.
Let \(S=\{v_1, v_2,\ldots, v_k\}\) be a subset of vectors in \(\R^n\text{.}\) Then show that \(L(S)\) is a vector subspace of \(\R^n\text{.}\)
Solution.
Let \(x,y\in L(S)\text{,}\) then by definitions there exist scalars \(\alpha_1,\ldots,\alpha_k \in \R\) such that \(x=\alpha_1v_1+\alpha_2v_2+\cdots+\alpha_kv_k\) and there exist another set of scalars \(\beta_1,\ldots,\beta_k \in \R\) such that \(y=\beta_1v_1+\beta_2v_2+\cdots+\beta_kv_k\text{.}\) Hence
\begin{align*}
x+y =\amp (\alpha_1v_1+\cdots+\alpha_kv_k)+\beta_1v_1+\cdots+\beta_kv_k\\
=\amp (\alpha_1+\beta_1)v_1+\cdots (\alpha_k+\beta_k)v_k\in L(S)
\end{align*}
The readers should understand what the properties of vector addition and scalar multiplications are used above.
Next, let \(\gamma\in \R\) be a scalar, then
\begin{equation*}
\gamma\cdot x=\gamma(\alpha_1v_1+\cdots+\alpha_kv_k)=(\gamma\alpha_1)v_1+\cdots+(\gamma\alpha_k)\in L(S)\text{.}
\end{equation*}
This shows that
\(L(S)\) is a vector subspace of
\(\R^n\text{.}\)
Checkpoint 2.2.3.
Try to prove the following.
If \(v\in \R^n\) is a non-zero vector in \(\R^n\text{,}\) then \(L(\{v\})=\{\alpha v:\alpha\in \R\}\) is a the line passing through origin and \(v\text{.}\) We shall denote \(L(v)\) by \(\R v\text{.}\)
Let \(S =\{e_1=(1,0),e_2=(0,1)\}\text{.}\) Then \(L(S)=\R^2\text{.}\)
Let \(S=\{(1,-1),(2,1)\}\text{,}\) then \(L(S)=\R^2\text{.}\)
Let \(S=\{(1,0,0),(0,1,0)\}\text{,}\) then \(L(S)=\{(x,y,0):x,y\in \R\}\) is the \(xy\)-plane given by the equation \(z=0\text{.}\)
Let \(S=\{(1,0,0),(0,1,0),(0,0,1)\}\text{,}\) then \(L(S)=\R^3\text{.}\)
\(S=\{(1,1,-1),(2,-1,3)\}\text{.}\) Then \(L(S)\) is a plane passing through the origin and the two points \((1,1,-1),(2,-1,3)\text{.}\) Can you find the scalars \(a,b,c\) such that \(L(S)\) is the plane represented by the equation \(ax+by+cz=0\text{?}\)
Let \(S=\{(1,1,-1),(1,2,3),(3,2,1)\}\text{.}\) Then \(L(S)=\R^3\text{.}\)
Solution.
Solution of all the problmes listed are easy verifications. Let us write a detailed proof of the last problem.
We need to show that \(L(S)=\R^3\text{,}\) that is, \(L(S)\subset \R^3\) and \(\R^3\subset L(S)\text{.}\) Let us define \(v_1:=(1,1,-1),v_2:=(1,2,3),v_3:=(3,2,1)\) for convenience. Clearly by, definition, \(L(S)\subset \R^3\text{.}\) To show \(\R^3\subset L(S)\text{,}\) we let \((a,b,c)\in \R^3\text{.}\) We need to find scalars, say \(x_1,x_2,x_3\in \R\) such that \((a,b,c)=x_1v_1+x_2v_2+x_3v_3\text{.}\) As a vector,
\begin{align*}
\begin{pmatrix} a\\b\\c\end{pmatrix} =\amp x_1\begin{pmatrix} 1\\1\\-1\end{pmatrix}+
x_2\begin{pmatrix} 1\\2\\3\end{pmatrix}+x_2\begin{pmatrix} 3\\2\\1\end{pmatrix}\\
=\amp \begin{pmatrix} 1 \amp 1 \amp 3\\1 \amp 2 \amp 2\\-1 \amp 3 \amp 1\end{pmatrix}\begin{pmatrix} x_1\\x_2\\x_3\end{pmatrix}=Ax\text{.}
\end{align*}
It is easy to check that the above system has a unique solution as rank of \(A\) is 3 (This can be easily verified by Sage). Hence \((a,b,c>\in \R^3\) and hence \(\R^3\subset L(S)\text{.}\) Hence we have \(L(S)=\R^3\text{.}\)
Example 2.2.4. Matrix Spaces.
Let \(A\) be a \(m\times n\) real matrix. For any vector \(x\in \R^n\text{,}\) \(Ax\in \R^m\text{.}\) Consider the following subsets
\({\cal N}(A):=\{x\in R^n:Ax=0\}\text{.}\) It is easy to check that \({\cal N}(A)\) is a subspace of \(\R^n\) called the null space of \(A\) or kernel of \(A\text{.}\)
\({\cal R}(A):=\{Ax:x\in \R^n\}\subset \R^m\) is a subspace of \(\R^m\text{,}\) called the image space or range space or column spaceof \(A\text{.}\) It is easy to see that for \(x\in \R^n\text{,}\) \(Ax\) is linear combinations of columns of \(A\text{.}\) We also denote \({\cal R}(A)\) by \({\rm col}(A)\text{.}\)
\({\cal L}(A):=\{y\in \R^m: A^Ty=0\}\) is a subspace of \(\R^m\) called the left null space of \(A\text{.}\)
Suppose we write rows of \(A\) as \(r_1,\ldots, r_m\text{.}\) Then each \(r_i\) is a vector in \(\R^n\text{.}\) Then linear span of \(\{r_1,\ldots, r_m\}\) is called the row space of \(A\) denoted by \({\rm row}(A)\text{.}\) It turns out that \({\rm row}(A)\) is a subspace of \(\R^n\text{.}\)
The four subspaces, namely \({\cal N}(A)\text{,}\) \({\rm col}(A)\text{,}\) \({\cal L}(A)\) and \({\rm row}(A)\) are called the fundamental subspaces associated with a matrix \(A\text{.}\)
