Show that $\langle u_1, u_2, u_3\rangle \subsetneq \langle v_1,v_2,v_3\rangle$ for the given vectors

Question

$$u_1 = (1,,1,2)\\ u_2 = (0,0,1) \\ u_3=(-1,-1,-1)$$

$$v_1=(1,1,1)\\ v_2=(1,1,0) \\ v_3 = (1,0,0)$$

To do this I have to prove that any vector $u_i$ can be written as a linear combination of the vectors in the V subspace so what I tried to do first was to find the V and U subspaces:

$$\langle v_1,v_2,v_3\rangle = \alpha_1(1,1,1)+\alpha_2(1,1,0)+\alpha_3(1,0,0)$$ $$\left\{ \begin{array}{c} \alpha_1+\alpha_2+\alpha_3=a \\ \alpha_1+\alpha_2=b\\ \alpha_1=c \end{array} \right.$$

I put this in a matrix and what I got was that the systen is possible for all values of a,b and c in $\mathbb{R}$: $$V = \{ (a,b,c) \in \mathbb{R}^3 : a,b,c \in \mathbb{R} \}$$

I did the same thing for subspace U and got $$U = \{ (a,b,c) \in \mathbb{R}^3 : a=b \}$$

How do I write $(a,a,c)$ as a linear combination of the vectors of V (now I realize that defining that subspace may have been unnecessary...)? I tried to do it this way:

$$(a,a,c) = \alpha_1(1,1,1)+\alpha_2(1,1,0)+\alpha_3(1,0,0)$$ $$\left\{ \begin{array}{c} \alpha_1+\alpha_2+\alpha_3=a \\ \alpha_1+\alpha_2=a\\ \alpha_1=c \end{array} \right. \\ \Leftrightarrow \\ (...) \\ \Leftrightarrow \left\{ \begin{array}{c} \alpha_1=c \\ \alpha_1+\alpha_2=a\\ \alpha_3=0 \end{array} \right.$$

I tried to solve this system using a matrix and got that it was possible for all values of $\mathbb{R}$ so I don't know. How do I do this?

Answer 1

The elements of the set $\{u_1,u_2,u_3\}$ aren't linearly independent, but $\{v_1,v_2,v_3\}$ it is. Since this last set genearate $\mathbb R^3$ and the first set generate a subspace of $\mathbb R^3$ then the claim of the tittle is true.

Answer 2

Try finding dimensions, $\dim(U)$, $\dim(V)$. You will find that

$\dim(U)=2$, and $\dim(V)=3$

So then it becomes obvious why subspaces aren't equal.