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

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

Let's say $\displaystyle u_{1}, u_{2}..., u_{k}$ are linearly independent vectors in $\displaystyle R^n$.

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

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

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

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

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

Show that $\displaystyle u_{1}, u_{2}..., u_{k}, u_{k+1}$ 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 $\displaystyle a_1u_{1}+a_2u_{2}...+a_ku_{k}+a_{k+1}u_{k+1}=0$

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

What can we conclude finally?

3. Thanks.

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

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

4. Originally Posted by owq
Thanks.

Can we say that since $\displaystyle a_1u_{1}+a_2u_{2}...+a_ku_{k}$ not equals $\displaystyle a_{k+1}u_{k+1}$, the equation will never be equal to zero unless$\displaystyle a_{k+1}=0$?
I am also new to the subject. Though what you say is correct in essence a more 'rigorous' argument will be
If $\displaystyle a_{k+1}$ is not equal to 0, it has an inverse $\displaystyle a_{k+1}^{-1}$. If we multiply the entire equation $\displaystyle a_1u_{1}+a_2u_{2}...+a_ku_{k}+a_{k+1}u_{k+1}=0$
by $\displaystyle a_{k+1}^{-1}$ we get$\displaystyle b_1u_{1}+b_2u_{2}...+b_ku_{k}+u_{k+1}=0$

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

Does this work: since $\displaystyle u_{k+1}$ and span($\displaystyle u_{1}, u_{2}..., u_{k}$) are linearly independent, and $\displaystyle u_{1}, u_{2}..., u_{k}$ are linearly independent, thus $\displaystyle u_{1}, u_{2}..., u_{k}, u_{k+1}$ are linearly independent?
Mu question would be - How do you define (and then prove) independence of a vector with a set span($\displaystyle u_{1}, u_{2}..., u_{k}$)?

5. If $\displaystyle a_{k+1}$ is not equal to 0, it has an inverse $\displaystyle a_{k+1}^{-1}$.
If we multiply the entire equation $\displaystyle a_1u_{1}+a_2u_{2}...+a_ku_{k}+a_{k+1}u_{k+1}=0$ by $\displaystyle a_{k+1}^{-1}$ we get

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

I see. Is there any other way to solve this?

6. 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