# Span of linear dependent set

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!

• How is it that a vector ($v_j$) is equal to a set of vectors (the span on the right-hand side)? – amd Jan 13 at 20:27

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 $$\{x_1,x_2,\ldots,x_{n-1}\}\subset Y$$.
• 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 $$\{v_1,\ldots v_{j-1},v_{j+1},\ldots,v_n\}$$,$$\operatorname{span}\bigl(\{v_1,\ldots v_{j-1},v_{j+1},\ldots,v_n\}\bigr)\supset\operatorname{span}\bigl(\{v_1,\ldots,v_n\}\bigr).$$