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?

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

1 Answer 1


Translating your argument into this form is totally possible, but the underlying reasoning is very much the same.

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.


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