Linear Dependent Vectors: v_1,v_2,...v_k,v Construct Linear Combination

[annoymage]

Homework Statement

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

Homework Equations

n/a

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"

 

[hunt_mat]

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

 

[annoymage]

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

 

[hunt_mat]

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

 

[annoymage]

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

 

[hunt_mat]

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

 

[annoymage]

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

not necessary, <br /> \lambda_{k+1}<br /> is non zero,

T_T I'm confused

 

[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

 

[annoymage]

you mean this right?

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

if yes, please check my argument here

<br /> <br /> \lambda_{1}v_{1}+\cdots +\lambda_{k}v_{k}=0<br /> <br /> are linear independent so, <br /> \lambda_{i}=0<br />

hence

<br /> <br /> 0v_{1}+\cdots +0v_{k}+\lambda_{k+1}v=0<br /> <br />

if <br /> \lambda_{k+1}=0<br /> then the vectors are linear independent, so <br /> \lambda_{k+1}\neq 0<br />

is this correct?

 

[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

 

[annoymage]

that crucial point is realllllllyyyy helpful, thankssssssss, btw it's <br /> v<br />

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