# Proof: Addition of vector makes the set remain linearly independent

• Oct 4th 2009, 09:49 AM
owq
Proof: Addition of vector makes the set remain linearly independent
Let's say $u_{1}, u_{2}..., u_{k}$ are linearly independent vectors in $R^n$.

Suppose $u_{k+1}$ is a vector in $R^n$ and $u_{k+1}$ is not a linear combination of $u_{1}, u_{2}..., u_{k}$.

Show that $u_{1}, u_{2}..., u_{k}, u_{k+1}$ are linearly independent.

I have some idea, but I can't think of a rigorous proof. Any ideas?
• Oct 4th 2009, 10:03 AM
aman_cc
Quote:

Originally Posted by owq
Let's say $u_{1}, u_{2}..., u_{k}$ are linearly independent vectors in $R^n$.

Suppose $u_{k+1}$ is a vector in $R^n$ and $u_{k+1}$ is not a linear combination of $u_{1}, u_{2}..., u_{k}$.

Show that $u_{1}, u_{2}..., u_{k}, u_{k+1}$ are linearly independent.

I have some idea, but I can't think of a rigorous proof. Any ideas?

I will try to get you started.
Let $a_1u_{1}+a_2u_{2}...+a_ku_{k}+a_{k+1}u_{k+1}=0$

Now we can have two cases:
1. $a_{k+1}$ not equal to $0$
Prove that this is not possible.
So we are left with
2. $a_{k+1}=0$. Under this what can you say about $a_1,a_2,...,a_k$?

What can we conclude finally?
• Oct 4th 2009, 10:31 AM
owq
Thanks.

Can we say that since $a_1u_{1}+a_2u_{2}...+a_ku_{k}$ not equals $a_{k+1}u_{k+1}$, the equation will never be equal to zero unless $a_{k+1}=0$?

Does this work: since $u_{k+1}$ and span( $u_{1}, u_{2}..., u_{k}$) are linearly independent, and $u_{1}, u_{2}..., u_{k}$ are linearly independent, thus $u_{1}, u_{2}..., u_{k}, u_{k+1}$ are linearly independent?
• Oct 4th 2009, 10:53 AM
aman_cc
Quote:

Originally Posted by owq
Thanks.

Can we say that since $a_1u_{1}+a_2u_{2}...+a_ku_{k}$ not equals $a_{k+1}u_{k+1}$, the equation will never be equal to zero unless $a_{k+1}=0$?

I am also new to the subject. Though what you say is correct in essence a more 'rigorous' argument will be
If $a_{k+1}$ is not equal to 0, it has an inverse $a_{k+1}^{-1}$. If we multiply the entire equation $a_1u_{1}+a_2u_{2}...+a_ku_{k}+a_{k+1}u_{k+1}=0$
by $a_{k+1}^{-1}$ we get $
b_1u_{1}+b_2u_{2}...+b_ku_{k}+u_{k+1}=0$

Which is not possible - hence $a_{k+1}$ is not equal to 0.

Quote:

Does this work: since $u_{k+1}$ and span( $u_{1}, u_{2}..., u_{k}$) are linearly independent, and $u_{1}, u_{2}..., u_{k}$ are linearly independent, thus $u_{1}, u_{2}..., u_{k}, u_{k+1}$ are linearly independent?

Mu question would be - How do you define (and then prove) independence of a vector with a set span( $u_{1}, u_{2}..., u_{k}$)?
• Oct 5th 2009, 12:53 AM
owq
If $a_{k+1}$ is not equal to 0, it has an inverse $a_{k+1}^{-1}$.
If we multiply the entire equation $a_1u_{1}+a_2u_{2}...+a_ku_{k}+a_{k+1}u_{k+1}=0$ by $a_{k+1}^{-1}$ we get

$u_{k+1} = -a_{k+1}^{-1}a_1u_{1}+-a_{k+1}^{-1}a_2u_{2}...+-a_{k+1}^{-1}a_ku_{k}$ which is not possible.
Hence $a_{k+1} = 0$.

I see. Is there any other way to solve this?
• Oct 5th 2009, 01:10 AM
aman_cc
Not that I know of. I feel this one is pretty fine, as we have just relied on axioms of field and vector space. I guess, any other approach should use a similar argument(about the existence of inverse - as I think that is the key axiom used here). Thanks