Let \(V\) be the real vector space spanned by the rows of the matrix \[ A = \begin{bmatrix} 3 & 21 & 0 & 9 & 0 \\ 1 & 7 & -1 & -2 & -1 \\ 2 & 14 & 0 & 6 & 1 \\ 6 & 42 & -1 & 13 & 0 \end{bmatrix}. \]
Find a basis for \(V\).
Tell which vectors \((x_1, x_2, x_3, x_4, x_5)\) are elements of \(V\).
If \((x_1, x_2, x_3, x_4, x_5)\) is in \(V\) what are its coordinates in the basis chosen in part (a)?
Solution
Row reduction gives \[ \operatorname{rref}(A)= \begin{bmatrix} 1&7&0&3&0\\ 0&0&1&5&0\\ 0&0&0&0&1\\ 0&0&0&0&0 \end{bmatrix}. \] Therefore a basis for the row space \(V\) is \[ v_1=(1,7,0,3,0), \quad v_2=(0,0,1,5,0), \quad v_3=(0,0,0,0,1). \]
A vector \(x=(x_1,x_2,x_3,x_4,x_5)\) lies in \(V\) exactly when it can be written \[ x=a v_1+b v_2+c v_3. \] Equating coordinates gives \[ x_2=7x_1, \qquad x_4=3x_1+5x_3, \] with \(x_1,x_3,x_5\) arbitrary. Thus \[ V=\{x\in\RR^5:x_2=7x_1,\ x_4=3x_1+5x_3\}. \]
For \(x\in V\), the coefficients are visibly \[ a=x_1, \qquad b=x_3, \qquad c=x_5. \] Hence the coordinate vector in the ordered basis \((v_1,v_2,v_3)\) is \[ [x]_{(v_1,v_2,v_3)}= \begin{bmatrix}x_1\\x_3\\x_5\end{bmatrix}. \]