# Showing two linear spans are the same through set theory (not linear combinations)

Consider the following problem:

Let $$V$$ be a $$\mathbb{K}$$-vector space. Let $$v_1, v_2 \in V$$ and $$\lambda \in \mathbb{K}$$. Show that $$\langle v_1, v_2\rangle = \langle v_1, v_2 + \lambda v_1\rangle$$, where $$\langle S\rangle$$ denotes the linear span of $$S$$.

It seems straigthforward to prove this in the following manner:

Let $$S = \{v_1, v_2 \}, S' = \{v_1, v_2 + \lambda v_1 \}$$. The span of $$S'$$ is

\begin{align*} \{x_1 v_1 + x_2(v_2 + \lambda v_1) \mid x_i\in \mathbb{K}\} &= \{x_1 v_1 + x_2 v_2 + \lambda x_1 \mid x_i\in \mathbb{K}\} \\ &= \{(x_1 + \lambda x_2)v_1 + x_2v_2 \mid x_i \in \mathbb{K}\} \end{align*}

Since $$\delta := x_1 + \lambda x_2 \in \mathbb{K}$$, $$\{(x_1 + \lambda x_2)v_1 + x_2v_2\} = \{\delta v_1 + x_2 v_2\}$$ is by definition the linear span of $$S$$ $$\blacksquare$$.

Once this proof was concluded, I attempted (for the sake of practicing and learning) to prove the spans are the same via the definition

The span of a set $$S$$ of vectors in a vector space $$V$$ is the interesection of all subspaces $$W_1, ..., W_n$$ of $$V$$ that contain $$S$$

In other words, I attempted to do the excercise thinking of the two spans as intersections instead of linear combinations. However, I haven't found a way to do it. How can one apply this "set-oriented" definition of linear span to prove the linear spans above are equal?

• Show that the collection of subspaces that contain $v_1$ and $v_2$ coincides with the collection of subspaces that contain $v_1$ and $v_2+\lambda v_1$. Commented Dec 20, 2022 at 23:43

Let $$V$$ be any subspace containing $$S_1 = \{ v_1, v_2\}$$. Since $$V$$ is a subspace (hence closed under linear combinations) we also have $$v_2 + \lambda v_1 \in V$$. So the subspace $$V$$ also contains $$S_2 = \{v_1, v_2 + \lambda v_1 \}$$.
Okay, what have we shown? We have shown that every subspace containing $$S_1$$ also contains $$S_2$$. In particular, the intersection of all of these, i.e. the span of $$S_1$$, contains $$S_2$$. This implies that it contains the span of $$S_2$$, i.e. $$\langle S_1 \rangle \supseteq \langle S_2 \rangle.$$
Now let $$V$$ any subspace containing $$S_2$$, and reverse the argument above to show the opposite containment.