# Linear independent

1. Aug 15, 2010

### annoymage

1. The problem statement, all variables and given/known data

if v_1,v_2,...,v_k be k linear independent vector, and if

v_1,v_2,...v_k,v be k+1 linear dependent vector, then

v is the linear combination of v_1,v_2,...,v_k

2. Relevant equations

n/a

3. The attempt at a solution

some of my attempt,(direct proof)

v_1,v_2,...v_k,v be k+1 linear dependent vector then when we write v_1,v_2,...v_k,v as linear combination of 0, there exist some coefficient not all of them 0. and i still no idea how to relate it to "v_1,v_2,...,v_k be k linear independent vector"

2. Aug 15, 2010

### hunt_mat

You can write:
$$\lambda_{1}v_{1}+\cdots +\lambda_{k}v_{k}+\lambda_{k+1}v=0$$
As a linealy dependent vector, then write $$\lambda_{k+1}=-1$$ to find:
$$v=\lambda_{1}v_{1}+\cdots +\lambda_{k}v_{k}$$

3. Aug 15, 2010

### annoymage

i don't understand, to find "$$v=\lambda_{1}v_{1}+\cdots +\lambda_{k}v_{k}$$" ?

4. Aug 15, 2010

### hunt_mat

If the vectors are linearly dependent then the $$\lambda_{i}$$ are all nonzero.

5. Aug 15, 2010

### annoymage

aren't some of them are non zero? not necessary all right?

6. Aug 15, 2010

### hunt_mat

There will be $$\lambda_{i}$$ which are nonzero, this will make the vector $$v$$

7. Aug 15, 2010

### annoymage

$$\lambda_{1}v_{1}+\cdots +\lambda_{k}v_{k}+\lambda_{k+1}v=0$$ are linearly dependent, then some of $$\lambda_{i}$$ are non zero..

not necessary, $$\lambda_{k+1}$$ is non zero,

T_T i'm confused

8. Aug 15, 2010

### hunt_mat

If $$\lambda_{k+1}\neq 0$$ then the vector will become linearly independent and then $$\lambda_{i}=0$$ will become zero. so $$\lambda_{k+1}\neq 0$$

9. Aug 16, 2010

### annoymage

you mean this right?

If $$\lambda_{k+1}= 0$$ then the vector will become linearly independent and then $$\lambda_{i}=0$$ will become zero. so $$\lambda_{k+1}\neq 0$$

if yes, please check my argument here

$$\lambda_{1}v_{1}+\cdots +\lambda_{k}v_{k}=0$$ are linear independent so, $$\lambda_{i}=0$$

hence

$$0v_{1}+\cdots +0v_{k}+\lambda_{k+1}v=0$$

if $$\lambda_{k+1}=0$$ then the vectors are linear independent, so $$\lambda_{k+1}\neq 0$$

is this correct?

10. Aug 16, 2010

### HallsofIvy

You are given that $\{v_1, v_2, \cdot\cdot\cdot, v_k\}$ are independent so we cannot have $\lambda_1 v_1+ \lambda_2 v_2+ \cdot\cdot\cdot+ \lambda_k v_k= 0$ unless $\lambda_1= \lambda_2= \cdot\cdot\cdot= \lambda_k= 0$

But we are also given that $\{v_1, v_2, \cdot\cdot\cdot, v_k, v\}$ are dependent- there exist $\lambda_1, \lambda_2, \cdot\cdot\cdot, \lambda_k, \lambda$, not all 0, such that $\lambda_1 v_1+ \lambda_2v_2+ \cdot\cdot\cdot+ \lambda_k v_k+ \lambda v= 0$.

Now, here is the crucial point: if $\lambda= 0$ $v$ would not be in the equation and we would have $\lambda_1 v_1+ \lambda_2 v_2+ \cdot\cdot\cdot+ \lambda_k v_k= 0$ with not all of $\lambda_1, \lambda_2, \cdot\cdot\cdot, \lambda_k$ equal to 0- which cannot happen. Thus, we must have $\lambda$ not 0.

No we can rewrite $\lambda_1 v_1+ \lambda_2 v_2+ \cdot\cdot\cdot+ \lambda_k v_k+ \lambda v= 0$ as $-\lambda v= \lambda_1v_1+ \lambda_2v_2+ \cdot\cdot\cdot+ \lambda_k vk$ and because $\lambda\ne 0$, we can divide through by $-\lambda$:
$$v= -\frac{\lambda_1}{\lambda}v_1- \frac{\lambda_2}{\lambda}v_2- \cdot\cdot\cdot- \frac{\lambda_k}{\lambda}v_k$$

Last edited by a moderator: Aug 16, 2010
11. Aug 16, 2010

### annoymage

that crucial point is realllllllyyyy helpful, thankssssssss, btw it's $v$

and thanks for all 3 quick reply, i'm still scrutinizing the other two, anyway, thanks again