Why is $L(u,v)$ the smallest subspace containing $u$ and $v$, where $L(S)$ is the linear span of $S \subset V$? In the lecture notes provided by my instructor, he proves for the statement $$L(\alpha,\beta)=L(u,v).$$
Statement: Let $V$ be a vector space over $F$ and $\alpha,\beta \in V$. Let A be a non-singular matrix as $$A=\begin{pmatrix}
    a & b\\
    c & d \\ 
    \end{pmatrix}.$$ Define $u=a\alpha+b\beta,v=c\alpha+d\beta.$ Then we claim that $$L(\alpha,\beta)=L(u,v).$$
Proof:
Here $u,v\in L(\alpha,\beta)$ and $L(u,v)$ is the smallest subspace of $V$ containing $u,v$. So, $L(u,v)\subset L(\alpha,\beta)$.
$$.....$$
Doubt:I have a problem with this statement as I am not able to figure the logic that how $L(u,v)$ is the smallest subspace of $V$. Please explain this logic.
Thanks for your time.
 A: For a set $S = \{ v_1,\dotsc,v_m\} \subset V$, $L(S)$ denotes the linear span of $S$, defined by
$$
L(S) = \{ c_1 v_1 + \dotsb + c_m v_m : c_i \in F \text{ for all } 1 \leq i \leq m \}.
$$
For any set $S \subset V$, $L(S)$ is in fact a subspace of $V$. (One can see this by checking that $L(S)$ is always non-empty, and is closed under addition and scalar multiplication.)

So, consider the vectors $u,v \in V$ as you have in your proof. What is $L(u,v)$? By definition, we have
$$
L(u,v) = \{ cu + dv : c,d \in F \}.
$$
In particular, when $c = 1$ and $d = 0$, we get that $u \in L(u,v)$, and when $c = 0$ and $d = 1$, we get that $v \in L(u,v)$. So, $L(u,v)$ is a vector subspace of $V$ that contains $u$ and $v$.
Now, suppose that $W \subset V$ is some subspace of $V$ that contains $u$ and $v$. We would like to show that $W$ contains all of $L(u,v)$. If we can do this, then it will imply that $L(u,v)$ is the smallest subspace of $V$ that contains both $u$ and $v$, in the above sense.
Well, since $u$ and $v$ belong to $W$, any linear combination of $u$ and $v$ also belongs to $W$. This is because $W$ is closed under addition and scalar multiplication, since it is a subspace of $V$. Hence, every vector of the form $cu + dv$ belongs to $W$ for arbitrary $c,d \in F$. But these are precisely all the vectors in $L(u,v)$. Hence, $W \supset L(u,v)$.
Hence, $L(u,v)$ is the smallest subspace of $V$ containing $u$ and $v$. In general, the same method of proof will show you that the linear span $L(S)$ of the set $S \subset V$ is the smallest subspace of $V$ containing $S$.
