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Let $\{v_1 \ldots v_n \}$ be a linear dependent set of vectors from a vector space $V$. We can show then that:

$$ v_j = span\{v_1 \ldots v_{j-1}, v_{j+1}, \ldots v_n \} $$

Because since it is linearly dependent, we can always choose a non-null (all together) set of elements $x_i$ of a field $\mathbf{F}$ such that:

$$ v_j = - \frac{\sum_{i\neq j}^{n} x_i v_i}{x_j}$$

Yet I have to prove a corollary which states that under the same conditions:

$$ span(v_1 \ldots v_n) = span(v_1 \ldots v_{j-1}, v_{j+1}, ,\ldots v_n) $$

and I am stuck because my results show that they are not the same. Any tips or advices to prove it? Thanks!

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  • $\begingroup$ How is it that a vector ($v_j$) is equal to a set of vectors (the span on the right-hand side)? $\endgroup$ – amd Jan 13 at 20:27
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Without loss of generality, we may assume $j=n$. Let $X=\text{span}\{x_1,x_2,\ldots,x_n\}$ and $Y=\text{span}\{x_1,x_2,\ldots,x_{n-1}\}$. It is easy to see that $$ X\geq Y $$from the definition of the spanned subspace. Now, to prove that $$ X\leq Y, $$ it is sufficient to show that $$ \{x_1,x_2,\ldots,x_{n}\}\subset Y. $$ But this is obvious from the fact that $x_n\in Y$ and $\{x_1,x_2,\ldots,x_{n-1}\}\subset Y$.

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  • $\begingroup$ Thank you! It has surely helped me :) $\endgroup$ – M.Gonzalez Jan 14 at 16:45
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Yes, they are the same:

  • since $\{v_1,\ldots v_{j-1},v_{j+1},\ldots,v_n\}\subset\{v_1,\ldots,v_n\}$,$$\operatorname{span}\bigl(\{v_1,\ldots v_{j-1},v_{j+1},\ldots,v_n\}\bigr)\subset\operatorname{span}\bigl(\{v_1,\ldots,v_n\}\bigr);$$
  • each element of $\{v_1,\ldots,v_n\}$ is a linear combination of elements of $\{v_1,\ldots v_{j-1},v_{j+1},\ldots,v_n\}$,$$\operatorname{span}\bigl(\{v_1,\ldots v_{j-1},v_{j+1},\ldots,v_n\}\bigr)\supset\operatorname{span}\bigl(\{v_1,\ldots,v_n\}\bigr).$$
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