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.


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

  • $\begingroup$ Can you add little details to it to explain why they are equal...? $\endgroup$ – Ritu Nov 4 '14 at 8:13
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    $\begingroup$ @Ritu, if $V$ is a subspace all the linear combinations of elements of $V$ are elements of $V$, so $\text{span}(V)=V$. $\endgroup$ – Martín-Blas Pérez Pinilla Nov 4 '14 at 8:31

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...

  • $\begingroup$ I wanted to know if it was right to think that spans are always linearly independent. It was getting mixed up.. So, this means the list of vectors may be both linearly independent and dependent? $\endgroup$ – Ritu Nov 4 '14 at 8:10
  • $\begingroup$ @Ritu Absolutely not. You can have $n$ linearly independent vectors $v_1,v_2,\dots, v_n$, but their span will not be linealry independent. $\endgroup$ – 5xum Nov 4 '14 at 8:14
  • $\begingroup$ Hmm..oky, I think I got it... Thank u..) $\endgroup$ – Ritu Nov 4 '14 at 8:16

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.


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