Is $span(v_1, . . . ,v_m)$ a linearly dependent or linearly independent set of vectors? Also, what will happen if we take span of span?

Question

We know that, the set of all linear combinations of $(v_1, . . . ,v_m)$ is called the span of $(v_1, . . . ,v_m)$, denoted by $span(v_1, . . . ,v_m)$. In other words, \begin{align} span(v_1, . . . ,v_m) =(a_1v_1 +· · ·+a_mv_m : a_1, . . . , a_m ∈ F)\end{align}

Also, the span of any list of vectors in V is a subspace of V and any list of vectors containing the $0$ vector is linearly dependent.

Now, since $span(v_1, . . . ,v_m)$ is a subspace of V, it must contain the additive identity, $0$ vector. So, does this mean that any of the vectors in $(v_1, . . . , v_m)$ may be zero? Or in other words, $(v_1, . . . , v_m)$ may be both linearly independent or linearly dependent?

Is $span(v_1, . . . ,v_m) = (a_1v_1 +· · ·+a_mv_m : a_1, . . . , a_m ∈ F)$ linearly independent or linearly dependent or both? Or since we cannot classify it in this respect?

I think since $span(v_1, . . . ,v_m)$ has linear combination of vectors, it itself must also be a set of vectors. What will happen if we take span of $span(v_1, . . . ,v_m)$?

Pretty confused!...Thanks in advance.

Answer 1

About the second question: as $\text{span}(v_1, . . . ,v_m)$ is a vector space $$\text{span}(\text{span}(v_1, . . . ,v_m))=\text{span}(v_1, . . . ,v_m).$$

Answer 2

The span of a set of vectors is a linear subspace, thus, it containts the zero vector, so it is never linearly independent. Your thinking is correct.

What is confuzing you? I mean, you didn't write any argument as to why the span would be linearly independent...

Answer 3

Let's use a concrete example. In $\mathbb R^3$ consider the set $\{e_1, e_2\}$ where $e_1 = (1,0,0)$ and $e_2 = (0,1,0)$. Then $$\text{span}\{e_1, e_2\} = \{\alpha_1 e_1 + \alpha_2 e_2 \mid \alpha_1, \alpha_2 \in \mathbb R \},$$ which is the $x$-$y$ plane.

Since every linear combination of elements in the $x$-$y$ plane is also a member of the $x$-$y$ plane,

$$\text{span}(\text{span}\{e_1, e_2\}) = \text{span} \{\alpha_1 e_1 + \alpha_2 e_2 \mid \alpha_1, \alpha_2 \in \mathbb R \} = \text{span}\{e_1, e_2\} $$

Your question about linear independence is a bit tricky to answer.

The set $\{ e_1, e_2 \}$ is a linearly independent set.

The set $\text{span}\{e_1, e_2\} = \{\alpha_1 e_1 + \alpha_2 e_2 \mid \alpha_1, \alpha_2 \in \mathbb R \}$ is an infinite set of vectors. It is a very linearly dependent set.