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Given the vectors $x_1,x_2,...,x_n$ and $y_1, y_2,...,y_k,$ where $n, k\in \Bbb N$ in a vector space $V$ over a field $\Bbb F$, prove that:

$$[\{x_1,x_2,...,x_n\}]=[\{y_1,y_2,...,y_k\}]$$

if and only if $$x_1, x_2,...,x_n \in [\{y_1,y_2,...,y_k\}]$$ $$y_1, y_2,...,y_n \in [\{x_1,x_2,...,x_n\}].$$

First question, what does $[\{x_1,x_2,...,x_n\}]=[\{y_1,y_2,...,y_k\}]$ really mean? How to think about it intuitively? $[\{x_1,x_2,...,x_n\}]$ means each vector in $V$ can be represented as a linear combination of $x$'s and $[\{y_1,y_2,...,y_k\}]$ means every vector in $V$ can be represented as a linear combination of $y$'s. But what does it mean they're equal?

Can I look at it as an equality of sets: two sets are equal if the first one is the subset of the second one and the second is the subset of the first one. But in that case, doesn't the proof follow form the definition?

The other method I tried was assuming the opposite:

$1)$ $[\{x_1,x_2,...,x_n\}]=[\{y_1,y_2,...,y_k\}]$ and $x_1, x_2,...,x_n \notin [\{y_1,y_2,...,y_k\}]$.

$2)$ $[\{x_1,x_2,...,x_n\}]=[\{y_1,y_2,...,y_k\}]$ and $y_1, y_2,...,y_n \notin [\{x_1,x_2,...,x_n\}].$

$3)$ $[\{x_1,x_2,...,x_n\}]\neq[\{y_1,y_2,...,y_k\}]$ and $x_1, x_2,...,x_n \in [\{y_1,y_2,...,y_k\}]$ and $y_1, y_2,...,y_n \in [\{x_1,x_2,...,x_n\}].$

But I don't know how to get a contradiction in each case, since I don't fully understand what $[\{x_1,x_2,...,x_n\}]=[\{y_1,y_2,...,y_k\}]$ means.

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    $\begingroup$ Does $[A] = \operatorname{span}A$? $\endgroup$ – Dan Rust Nov 22 '16 at 18:31
  • $\begingroup$ well $[\{x_1,x_2\}]$ is a vector subspace. Two vector subspaces are equals if their elements are indentical. $\endgroup$ – Hello Nov 22 '16 at 18:31
  • $\begingroup$ @DanRust Looks like yes $\endgroup$ – Hello Nov 22 '16 at 18:31
  • $\begingroup$ @DanRust Yes, it is the span. $\endgroup$ – lmc Nov 22 '16 at 18:32
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    $\begingroup$ In general, given a set $ S\subseteq V $, $[S] $ means the subspace generated by $S $. This is just the set of linear combinations of the form $\alpha v $ where $\alpha \in \mathbb{F} $ and $ v\in S$. $\endgroup$ – Xam Nov 22 '16 at 18:33
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Note that we always have $z_i \subset \operatorname{sp} \{ z_1,...,z_l\}$ for $i=1,...,l$, and if $ A \subset B$, then $\operatorname{sp} A \subset \operatorname{sp} B$.

Let $X = \operatorname{sp} \{ x_1,...,x_n \}$, $Y = \operatorname{sp} \{ y_1,...,y_k \}$.

Suppose $X = Y$, then $x_i \in X \subset Y$ and similarly for the $y_i$.

Now suppose $\{ x_1,...,x_n \} \subset Y$ and $\{ y_1,...,y_k \} \subset X$. Then we have $X = \operatorname{sp} \{ x_1,...,x_n \} \subset \operatorname{sp} Y = Y$, and similarly with the roles of $X,Y$ reversed. Hence $X=Y$.

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HINT: if $\{x_1,x_2,\ldots,x_n\}\subset{\rm span}\{y_1,y_2,\ldots,y_k\}$ then by definition of the span of a set the elements of the first set is a linear combination of the elements of the second.

Then every linear combination of the first set is in the span of the second (why?) This, together with the reversed inclusion of the other span means that their span are equal.

To prove it choose any vector of one span and show that it is in the other span.

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