Proof: Addition of vector makes the set remain linearly independent

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October 4th 2009, 08:49 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?
October 4th 2009, 09:03 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?
October 4th 2009, 09:31 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?
October 4th 2009, 09:53 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 $<br /> b_1u_{1}+b_2u_{2}...+b_ku_{k}+u_{k+1}=0$

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

(Any comments from other senior members here will be helpful.)

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( $u_{1}, u_{2}..., u_{k}$ )?
October 4th 2009, 11:53 PM
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?
October 5th 2009, 12:10 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