Loomis and Sternberg Problem 1.31

$\mathbf{a}=<1,0,-0.5>, \mathbf{b}=<1,1,0> \in M$. Let $L$ be the set $\{t_1\mathbf{a} + t_2\mathbf{b}: t_1, t_2 \in \mathbb{R}\}$. It’s easy to see that any $\beta \in L$ is also in $M$ so $L \subseteq M$. How do I show that any $\alpha \in M$ is also in L?

Thanks

From $x_1 - x_2 + 2 x_3 = 0 $, we choose $x_2 = t $ and $x_3 = s $, then $x_1 = t - 2 s $, i.e.

$(x_1, x_2, x_3) = t (1, 1, 0) + s (-2, 0, 1) = t \mathbf{a} + s \mathbf{b} $

with $\mathbf{a} = (1 , 1, 0) , \mathbf{b} = (-2, 0, 1) $

Now suppose $\mathbf{c} = t_1 \mathbf{a} + s_1 \mathbf{b} $ and $\mathbf{d} = t_2 \mathbf{a} + s_2 \mathbf{b} $

In matrix-vector form we have the following relation

$[\mathbf{c}, \mathbf{d} ] = [\mathbf{a}, \mathbf{b}] \begin{bmatrix} t_1 && t_2 \\ s_1 && s_2 \end{bmatrix} $

Assuming the $t_1 s_2 - t_2 s_1 \ne 0 $ , then $\mathbf{c}$ and $\mathbf{d} $ are linearly independent, because then $[\mathbf{c}, \mathbf{d}] \mathbf{\lambda} = \mathbf{0} $ would imply that $\mathbf{\lambda} = 0$.

Thus, in this case, we can invert the $2 \times 2$ matrix on the right hand side, and get

$[\mathbf{a}, \mathbf{b} ] = [\mathbf{c}, \mathbf{d}] \begin{bmatrix} t_1 && t_2 \\ s_1 && s_2 \end{bmatrix}^{-1} $

Now if vector $\mathbf{x}$ is a linear combination of $\mathbf{c}, \mathbf{d}$ , then obviously it is a linear combination of $\mathbf{a}, \mathbf{b}$ and is thus in $M$. And if $\mathbf{x} $ is in $M$, then it is a linear combination of $\mathbf{a}, \mathbf{b}$ and therefore a linear combination of $\mathbf{c}, \mathbf{d}$.

Hence, $\mathbf{x} \in M$ if and only if it is a linear combination of $\mathbf{c}, \mathbf{d} $.