1. ## Linear Independence/Span Proof

Suppose that V is a vector space, that $X=\{v_1,...v_n\} \subset V$ and that $w \in span X$. Set $Y=\{w, v_1,...,v_n\}$. If necessary you may denote the field of scalars by F.

(a) Show that $Y$ is linearly dependent.

(b) Show that $spanY = spanX$

This is my attempt for part (a): We know that since w is in the span of X then it's a linear combination of its vectors (i.e. $w=a_1 v_1 +...+a_n v_n$, and $a_1,...,a_n$ are scalars).

Suppose there are real numbers $\lambda_1, \lambda_2,...,\lambda_m$ (m=n+1) such that:

$\lambda_1 w+ \lambda_2 v_1 + \lambda_m v_n = 0$

Then $Y$ is linearly dependent if $\lambda_1 = \lambda_2 =...= \lambda_m = 0$. Now I'm not sure how to prove this.

Of course we can rewrite it as:

$\lambda_1 (a_1 v_1 +...+a_n v_n)+ \lambda_2 v_1 + \lambda_m v_n = 0$

But that doesn't seem to help...

2. Originally Posted by Roam
Suppose that V is a vector space, that $X=\{v_1,...v_n\} \subset V$ and that $w \in span X$. Set $Y=\{w, v_1,...,v_n\}$. If necessary you may denote the field of scalars by F.

(a) Show that $Y$ is linearly dependent.

(b) Show that $spanY = spanX$

This is my attempt for part (a): We know that since w is in the span of X then it's a linear combination of its vectors (i.e. $w=a_1 v_1 +...+a_n v_n$, and $a_1,...,a_n$ are scalars).

Suppose there are real numbers $\lambda_1, \lambda_2,...,\lambda_m$ (m=n+1) such that:

$\lambda_1 w+ \lambda_2 v_1 + \lambda_m v_n = 0$

Then $Y$ is linearly dependent if $\lambda_1 = \lambda_2 =...= \lambda_m = 0$. Now I'm not sure how to prove this.

Of course we can rewrite it as:

$\lambda_1 (a_1 v_1 +...+a_n v_n)+ \lambda_2 v_1 + \lambda_m v_n = 0$

But that doesn't seem to help...
$Y=\left\{v_1,v_2,...,v_m,w \right\}$ is linearly dependent if there exist some scalars $a_1,a_2,...,a_n \in F$, not all 0, such that $a_1v_1 + a_2v_2 + ... + a_mv_m + a_nw = 0$

Now, assume there exist some scalars $a_1,a_2,...,a_n \ s.t. \ a_1v_1 + a_2v_2 + ... + a_mv_m+a_nw = 0$. Recall that $w=\beta_1v_1 + \beta_2v_2 + ... + \beta_mv_m$ for some $\beta_1,...,\beta_m \in F$. If $w=0$, we are done. Assume $w \neq 0$.

Rewriting the equation, we get:

$a_1v_1 + a_2v_2 + ... + a_mv_m + a_n\beta_1v_1 + a_n\beta_2v_2 + ... + a_n\beta_mv_m = 0 \Rightarrow$ $(a_1+a_n\beta_1)v_1 + (a_2 + a_n\beta_2)v_2 + ... + (a_m + a_n\beta_m)v_m = 0$.

So, if we choose, say, $a_n=1$, then we'll choose $a_1 = -\beta_1, a_2 = -\beta_2,...,a_m = -\beta_m$, and now we have $a_1,a_2,...,a_n \in F$, not all 0, such that $a_1v_1 + a_2v_2 + ... + a_mv_m + a_nw = 0$ and thus Y is linearly dependent.

Can you follow on the second part from here?

3. Originally Posted by Roam
Suppose that V is a vector space, that $X=\{v_1,...v_n\} \subset V$ and that $w \in span X$. Set $Y=\{w, v_1,...,v_n\}$. If necessary you may denote the field of scalars by F.

(a) Show that $Y$ is linearly dependent.

(b) Show that $spanY = spanX$

This is my attempt for part (a): We know that since w is in the span of X then it's a linear combination of its vectors (i.e. $w=a_1 v_1 +...+a_n v_n$, and $a_1,...,a_n$ are scalars).

Suppose there are real numbers $\lambda_1, \lambda_2,...,\lambda_m$ (m=n+1) such that:

$\lambda_1 w+ \lambda_2 v_1 + \lambda_m v_n = 0$

Then $Y$ is linearly dependent if $\lambda_1 = \lambda_2 =...= \lambda_m = 0$. Now I'm not sure how to prove this.

Of course we can rewrite it as:

$\lambda_1 (a_1 v_1 +...+a_n v_n)+ \lambda_2 v_1 + \lambda_m v_n = 0$

But that doesn't seem to help...
There's your part a) ! You proved it but didn't realize.

Since $w=a_1 v_1 +...+a_n v_n$ you have $0=-w+a_1 v_1 +...+a_n v_n$ and you're done; $Y$ is linearly dependent!

4. Hi Defunkt! Thanks a lot, I get it now! Btw, I think it has to be $a_mv_n$ in your post. So it should be:
$(a_1+a_n\beta_1)v_1 + (a_2 + a_n\beta_2)v_2 + ... + (a_m + a_n\beta_m)v_n = 0$

Now, I don't know how to prove part (b)...

But here is my attempt: We can try showing that an element in Y belongs to the span of X, and an element in X belongs to the span of Y.

Since we know that $w \in spanX$, we also know that $w \in Y$.

Therefore $spanX \subset Y$ (?)

Now we can take $v_1 \in X$ and we know that v_1 belongs to the $spanY$,

Therefore $X \subset spanY$ (?)

I'm very confused...

5. I don't get part (b)

6. Let $x \in span(X)$, then $x = \alpha_1v_1 + \alpha_2v_2 + ... + \alpha_nv_n$ where $\alpha_1, \alpha_2,...,\alpha_n \in F$

If $x=0$ then obviously $x \in span(Y)$; assume $x \neq 0$.

We want to show that $x \in span(Y)$, ie. that there exist $\beta_1, \beta_2,...,\beta_n,\beta_{n+1} \in F$ such that $x = \beta_1v_1 + \beta_2v_2 + ... + \beta_nv_n + \beta_{n+1}w$.

It can easily be seen that if we let $\beta_i = \alpha_i \ \forall 1 \leq i \leq n, \ \beta_{n+1} = 0$ then $x \in span(Y)$. So $span(X) \subset span(Y)$

Now let $y \in span(Y)$. Then $y = \gamma_1v_1 + ... + \gamma_nv_n + \gamma_{n+1}w$ where $\gamma_1,\gamma_2,...,\gamma_n,\gamma_{n+1} \in F$.

We want to show that there exist $\lambda_1,\lambda_2,...,\lambda_n \in F$ such that $y = \lambda_1v_1 + \lambda_2v_2 + ... + \lambda_nv_n$.

But $y = \gamma_1v_1 + ... + \gamma_nv_n + \gamma_{n+1}w = \gamma_1v_1 + ... + \gamma_nv_n + \gamma_{n+1}\delta_1v_1 + \gamma_{n+1}\delta_2v_2 + ... + \gamma_{n+1}\delta_nv_n$
$= (\gamma_1+\gamma_{n+1}\delta_1)v_1 + (\gamma_2+\gamma_{n+1}\delta_2)v_2 + ... + (\gamma_n + \gamma_{n+1}\delta_n)v_n$

So if we take $\lambda_i = \gamma_i + \gamma_{n+1}\delta_i \ \forall 1 \leq i \leq n$, then $y = \lambda_1v_1 + \lambda_2v_2 + ... + \lambda_nv_n \Rightarrow y \in span(X)$

So $span(Y) \subset span(X)$ and $span(X) \subset span(Y) \Rightarrow span(X) = span(Y)$

7. ## lin dependence

Alright, so I'm still new to the whole lin. algebra thing w/ the span S =R^3

I'm trying to find an example of a subset S of R^3 that is also lin. dependent and span S = R^3....

I would have dim = 3, right?

but what about the values for the subset, can I use something S: {1,x,x^2?}

what does it mean to be lin. dependent?