# Yet another proof

let me know if u think i did this proof correctly...

use the definition of linear independence to prove that if $${u_1,u_2,...,u_n}$$ is a linearly independent set, then $${u_1,u_2,...,u_k}$$ is also a linearly independent set for any
$$k = 2,3,...,n-1$$

well heres what i did...
if the set of vectors $${u_1,u_2,...,u_n}$$ is linearly independent then none of the vectors in that set is linear combination of the others. therefore
$$c_1 u_1 + c_2 u_2 + ... + c_n u_n = \vect{0}$$ is satisfied only by the set of scalars $${c_1,c_2,...,c_n}$$ that is $$c_1 = 0, c_2 = 0, ..., c_n = 0$$ then.. $$c_1 u_1 + c_2 u_2 +...+c_(n-1) u_(n-1) = -c_n u_n$$ should only be satisfied by the same set where $$c_1 = 0, c_2 = 0, ..., c_n = 0$$ because no vector is a linear combination of the others by definition. thus this can be expanded to...$$c_1 u_1 + c_1 u_2 + ... + c_k u_k = -c_n u_n -c_(n-1) u_(n-1) - ... -c_(k+1) u_(k+1)$$ the only solution is the set where $$c_1 = 0, c_2 = 0, ..., c_n = 0$$, thus the set $${u_1,u_2, ...,u_k}$$ is linearly independent for any $$k=2,3,...,n-1.$$

tell me what u think..

Last edited:

matt grime
Homework Helper
I think that this question doesn't even need proof.

If there is a linear combination of u_1,..,u_k that is zero, then there is a linear combination of u_1,..,u_n (k<=n) that is also zero. done. You're writing far too much.

HallsofIvy
Homework Helper
Well, yes, it needs a proof, its just a trivial one!

Phymath: Suppose $$\{u_1,u_2,...,u_{n-1}\}$$ were not linearly independent. That would mean that there exist constants $${\kc_1,c_2,...,c_{n-1}\}$$, not all 0, such that $$c_1u_1+ c_2u_2+...+c_{n-1}u_{n-1}= 0$$.
What does that tell you about the orginal set $$\{u_1,u_2,...,u_n\}$$?

well idk i guess thats why i asked the question in the first place. I suppose it could mean that u_(n-1) is a linear combination of the procedding u vectors..$$u_1, u_2, ...,u_{n-2}$$ no? but i thought by defination that can't be...i suppose that i tells excaztly what i said in my proof that there are no scalar constants to be a solution to that set other then all the constants have to be 0. because taking ur equation if they are linearly dependent (aka not linearly independent) then u can do this, which is by defintion only solved if the constants are zero, please help me out here...i don't get ur point
$$c_1 u_1 +c_2 u_2 +...+c_(n-2) u_(n-2) = -c_(n-1) u_(n-1)$$

mathwonk
Homework Helper
2020 Award
linear dependence means zero is a non trivial linear combination of some of the vectors.

but if zero is a linear combination of some of the first k vectors, then zero is also (the same) linear combination of some of the full set of n vectors.

how's this?

mathwonk
Homework Helper
2020 Award
another point of view: linear independence means the map sending the standard unit vectors to those vectors is injective. but an injective map on n space restricts to an injective map on a k dimensional subspace.

ok ok i know that makes it worse.

mathwonk said:
linear dependence means zero is a non trivial linear combination of some of the vectors.

but if zero is a linear combination of some of the first k vectors, then zero is also (the same) linear combination of some of the full set of n vectors.

how's this?
yes but i have to work this backwards...n then to k not k then to n

mathwonk