Question

Let L = {(1, 1, 1, 1, −4),(1, −1, 3, −2, −1)}. Find 6 vectors in the collection,
say H, such that L ∪ H spans the entire space.

Answer 1

We can see that all vectors in $L\cup\{u_1,u_2,u_3,u_4,u_5,u_6,u_7,u_8\}$ satisfy the equation $x_1+x_2+x_3+x_4+x_5=0$.

Indeed, $l_1=(1,1,1,1,-4)$ , so 1+1+1+1+(-4)=0

$l_2=(1,-1,3,-2,-1)$ , so 1+(-1)+3+(-2)+(-1)=0

$u_1=(2,-3,4,-5,2)$ , 2+(-3)+4+(-5)+2=0

$u_2=(-6,9,-12,15,-6)$ , (-6)+9+(-12)+15+(-6)=0

$u_3=(3,−2,7,−9,1)$ , 3+(-2)+7+(-9)+1=0

$u_4=(2,−8,2,−2,6)$ , 2+(-8)+2+(-2)+6=0

$u_5=(−1,1,2,1,−3)$ , (-1)+1+2+1+(-3)=0

$u_6=(0,−3,−18,9,12)$ , 0+(-3)+(-18)+9+12=0

$u_7​=(1,0,−2,3,−2)$ , 1+0+(-2)+3+(-2)=0

$u_8​=(2,−1,1,−9,7)$ , 2+(-1)+1+(-9)+7=0

So all vectors in $L\cup\{u_1,u_2,u_3,u_4,u_5,u_6,u_7,u_8\}$ satisfy the equation $x_1+x_2+x_3+x_4+x_5=0$, that is $(1,1,1,1,1)\cdot (x_1,x_2,x_3,x_4,x_5)=0$

Let $T\vec{x}=(1,1,1,1,1)\cdot\vec{x}$ , then $L\cup\{u_1,u_2,u_3,u_4,u_5,u_6,u_7,u_8\}\subset\ker T$. Since $ImT\neq{0} (\bigl|(1,1,1,1,1)\bigr|^2\in ImT$) and $ImT\subset R$ , we have $\dim ImT=1$

Dimensional theorem (https://en.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces) tells us that $\dim\ker T+\dim ImT=\dim V$, that is $\dim\ker T+1=5$, so $\dim\ker T=4$.

Since $\langle L\cup\{u_1,u_2,u_3,u_4,u_5,u_6,u_7,u_8\}\rangle\subset\ker T$, we have $\dim\langle L\cup\{u_1,u_2,u_3,u_4,u_5,u_6,u_7,u_8\}\rangle\le\dim\ker T=4$, therefore $\langle L\cup\{u_1,u_2,u_3,u_4,u_5,u_6,u_7,u_8\}\rangle\neq V$ . We cannot choose $H$ such that $\langle L\cup H\rangle = V$.