Math 1220 Linear Algebra

Subsubsection 5.9.4.1 Subspace examples

• Let \(\vec x\in\R^n\) and \(S=\{\vec y\mid \vec y=r\vec x, r\in\R\}\)

  1. If \(\vec y_1\) and \(\vec y_2\) are in \(S\text{,}\) then \(\vec y_1=r_1\vec x\) and \(\vec y_2=r_2\vec x\text{,}\) and so \(\vec y_1 + \vec y_2=r_1\vec x + r_2\vec x = (r_1+r_2)\vec x\in S\text{.}\)

  2. If \(\vec y\in S\) and \(r\in \R\text{,}\) then \(\vec y=s\vec x\) for some \(s\in\R\text{,}\) and \(r\vec y=r(s\vec x) = (rs)\vec x\in S\text{.}\)

• \(S=\{\vec0\}\text{.}\) This follows since \(\vec0+\vec0 = \vec0\) and \(r\vec0 = \vec0\text{.}\)

• Let \(A\) be an \(m\times n\) matrix and let \(S=\{\vec x\in\R^n \mid A\vec x=\vec0\}\text{.}\)

  1. If \(\vec x_1\) and \(\vec x_2\) are in \(S\text{,}\) then \(A\vec x_1=\vec0\) and \(A\vec x_2=\vec 0\text{,}\) and so \(A(\vec x_1+ \vec x_2)=A\vec x_1 + A\vec x_2 =\vec0+\vec0 = \vec0\text{.}\) Thus \(\vec x_1+\vec x_2\in S\text{.}\)

  2. If \(\vec x\in S\) and \(r\in \R\text{,}\) then \(A\vec x=\vec 0\text{,}\) and \(A(r\vec x)=r(A\vec x) = r\vec0=\vec0\text{.}\) Hence \(r\vec x\in S\text{.}\)

  This subspace is called the null space of \(A\text{.}\)

• Let \(H\) be hyperplane in \(\R^n\) containing \(\vec0\text{,}\) that is, a set of vectors \(\vec x=(x_1,\ldots,x_n)\) satisfying an equation

  \begin{equation*} a_1x_1+a_2x_2+\cdots+a_nx_n=0 \end{equation*}

  where at least one \(a_i\not=0\text{.}\)

  1. If \(\vec x=(x_1,\ldots,x_n)\) and \(\vec y=(y_1,\ldots,y_n)\) are in \(H\text{,}\) then from the equations

     \begin{gather*} a_1x_1+a_2x_2+\cdots+a_nx_n=0,\\ a_1y_1+a_2y_2+\cdots+a_ny_n=0, \text{ and}\\ \vec x+\vec y=(x_1+y_1,\ldots,x_n+y_n) \end{gather*}

     it follows that

     \begin{align*} a_1(x_1+y_1)+\cdots+a_n(x_n+y_n)\amp = (a_1x_1+\cdots+a_nx_n) + (a_1y_1+\cdots+a_ny_n)\\ \amp=0+0\\ \amp=0, \end{align*}

     and \(\vec x+\vec y\in H\text{.}\)

  2. \(r\vec x=(rx_1,\ldots,rx_n)\text{,}\) and

     \begin{equation*} a_1(rx_1)+\cdots+a_n(rx_n)=r(a_1x_1+\cdots+a_nx_n)=r0=0 \end{equation*}

     and so \(r\vec x\in H\text{.}\)

• Let \(\mathcal{X}=\{\vec x_1, \vec x_2,\ldots, \vec x_m\}\) be a set of vectors in \(\R^n\text{,}\) and let

  \begin{equation*} S=\Span\mathcal{X}= \{r_1\vec x_1 + r_2\vec x_2+\cdots+ r_m\vec x_m \mid r_i\in\R\}. \end{equation*}

  1. Let \(\vec y_1\) and \(\vec y_2\) be in \(S\text{,}\) that is

     \begin{gather*} \vec y_1=r_1\vec x_1 + r_2\vec x_2+\cdots+ r_m\vec x_m\\ \vec y_2=s_1\vec x_1 + s_2\vec x_2+\cdots+ s_m\vec x_m. \end{gather*}

     Then

     \begin{equation*} \vec y_1+\vec y_2= (r_1+s_1)\vec x_1 + (r_2+s_2)\vec x_2+\cdots+ (r_m+s_m)\vec x_m \in S. \end{equation*}

  2. If \(\vec y=r_1\vec x_1 + r_2\vec x_2+\cdots+ r_m\vec x_m\text{,}\) then

     \begin{align*} r\vec y\amp=r(r_1\vec x_1 + r_2\vec x_2+\cdots+ r_m\vec x_m)\\ \amp=(rr_1)\vec x_1 + (rr_2)\vec x_2+\cdots+ (rr_m)\vec x_m \in S. \end{align*}

• If \(A\) is an \(m\times n\) matrix with rows \(R_1, R_2,\ldots,R_m\) and columns \(C_1,C_2,\ldots,C_n\text{.}\) Then

  • The row space of \(A\) is \(\Span \{R_1,R_2,\ldots R_m\}\subseteq\R^n\text{.}\)

  • The column space of \(A\) is \(\Span \{C_1,C_2,\ldots C_n\}\subseteq\R^m\text{.}\)

• If \(S_1\) and \(S_2\) are subspaces of \(\R^n\text{,}\) then so is \(S_1\cap S_2\text{.}\)

  1. Let \(\vec x\in S_1\cap S_2\) and \(\vec y\in S_1\cap S_2\text{.}\) Then \(\vec x\in S_1\) and \(\vec y\in S_1\text{.}\) Since \(S_1\) is a subspace, \(\vec x+\vec y\in S_1\text{.}\) Similarly \(\vec x+\vec y\in S_2\text{,}\) and so \(\vec x+\vec y\in S_1\cap S_2\text{.}\)

  2. Let \(\vec x\in S_1\cap S_2\) and \(r\in\R\text{.}\) Then \(\vec x\in S_1\) and \(S_1\) a subspace implies \(r\vec x\in S_1\text{.}\) Similarly, \(r\vec x\in S_2\text{,}\) and so \(r\vec x\in S_1\cap S_2\text{.}\)

Subsubsection 5.9.4.2 Bases for a subspaces