Two lines in $\mathbb R^{3}$ which contain 0 when summed make a proper subspace Let $L_{1}$ and $L_{2}$ be lines in $\mathbb R^{3}$ which contain 0. Show that $L_{1}$ + $L_{2}$ is a proper subspace of $\mathbb R^{3}$. Can you describe the subspace as the solution set of a system of linear equations. I am stuck, I can't figure out how to prove this. Could someone help with this?
Thanks!
 A: Let $$L_1 = \{ (a_1t, b_1t, c_1t):t \in \mathbb{R}\}$$ $$L_2 = \{ (a_2s, b_2s, c_2s):s \in \mathbb{R}\}$$ be the two lines Spanned by the vectors $(a_1, b_1, c_1), (a_2, b_2, c_2)$. Then
$$L_1+L_2 = \{ (a_1t+a_2s, b_1t+b_2s, c_1t+c_2s):t,s \in \mathbb{R}\} $$
is a plane of dimension 2, so it cannot be the whole $\mathbb{R}^3$. 
Since $(a_1, b_1, c_1), (a_2, b_2, c_2)$ are linearly independent, you have that one among
$$a_1b_2-a_2b_1, a_1c_2-a_2c_1, c_1b_2-c_2b_1$$
is non zero (otherwise...?). 
Then, an equation (which is non trivial) satisfied by all points of $L_1+L_2$ is
$$(b_2c_1-b_1c_2) x + (a_1b_2-a_2b_1) y + (b_1a_2-a_1b_2)z = 0$$
A: Let $L_1 = t\left<x_1,x_2,x_3\right>$, $L_2 = s\left<y_1,y_2,y_3\right>$, (i.e. $L_1$ goes through the origin and  $x = \left<x_1,x_2,x_3\right>$, $L_2$ does likewise for $y$). 
Then the subset spanned by $L_1$ and $L_2$ is
$$
S = \{\left<b_1,b_2,b_3\right> \vert \,\exists s,t : b_1 = x_1t + y_1s, b_2 = x_2t + y_2s, b_3 = x_3t + y_3s\}.
$$
Can you show this is a subspace? In other words, that it is closed under linear combinations? (if $b,c \in S$, need to show $\alpha b + \beta c \in S$ by demonstrating the existence of appropriate $s,t$; write out the equation in components, and then this isn't too hard).
To write as a system, note that the condition is that there exist $s,t$ so that
$$
\begin{pmatrix} x_1 & y_1 \\x_2 & y_2 \\x_3 & y_3 \\ \end{pmatrix}\begin{pmatrix}s \\t \end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ b_3\end{pmatrix}.
$$
Assuming by "proper" subspace, you mean it is not all of $\mathbb R^3$, can you now see why this must be the case? That is, is there any way that the map above is surjective (i.e. for fixed $x,y$, is there any way a solution $s,t$ exists for all $b \in \Bbb R^3$?)
A: Since the lines contain zero, we can write $L_k = \{ t d_k | t \in \mathbb{R} \}$
for some vectors $d_k \in \mathbb{R}^3$.
Then $L_1+L_2 = \{ t_1 d_1+t_2 d_2 | t_k \in \mathbb{R} \} = {\cal R} A$,
where $A= \begin{bmatrix}d_1 & d_2\end{bmatrix}$.
If $L_1, L_2$ are the same line, then find $b_2,b_3$ so that
$d_1,b_2,b_3$ form an orthonormal basis. Then you can write
$L_1 = L_1+L_2 = \{ x | b_2^T x =0 , b_3^T x = 0 \}$.
If $L_1,L_2$ are different, then find $b_3$ so that
$d_1,d_2,b_3$ form an orthonormal basis. Then you can write
$L_1 = L_1+L_2 = \{ x |  b_3^T x = 0 \}$.
A: Since any line through the origin in $\Bbb R^3$ is generated by a non-zero vector, so assume that $L_1$ and $L_2$ are distinct lines and say $L_1=\text{Sp}\{b_1\}$ and $L_2=\text{Sp}\{b_2\}$. Clearly then $b_1$ and $b_2$ are linearly independent (as $L_i$'s are assumed to be distinct lines through origin) and $L_1+L_2=\text{Sp}\{b_1,b_2\}$. So $L_1+L_2$ has dimension $2$, which is a proper subspace as $\dim \Bbb R^3$ is $3$.
For the other part, first extend  the linearly independent subset $\{b_1,b_2\}$ to a basis, say, $B=\{b_1,b_2,b_3\}$ for $\Bbb R^3$. Define a linear map $T:\Bbb R^3\to \Bbb R^3$ by setting its action on the basis vectors i.e. $T(b_1)=0=T(b_2)$ and $T(b_3)=b_3$. Clearly $\ker(T)=\text{Sp}\{b_1,b_2\}=L_1+L_2$. Also since every linear map $T:\Bbb R^3\to \Bbb R^3$ can be expressed as  $T(x)=Ax$, for some $A\in M_3(\Bbb R)$ and $A$ can be uniquely determined by the image of the basis vectors under $T$ (i.e. filling up the first column of $A$ by the co-ordinate vectors of $T(b_1)$ and so on). So $L_1+L_2$ is the kernel of $T$ which is in fact equivalent to saying that $L_1+L_2$ is the solution space of  the system of equation given by $Ax=0$ (since $x\in \ker(T)\Longleftrightarrow 0=T(x)= Ax=0$).
