1. ## Basis Proof

Let ${u,w,v}$ be a basis for vector space V. Prove that ${u+v+w, v+w,w}$ is also a basis for vector space V
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Ok so the show that this is a basis, it must satisfy being linearly independent and must be able to span any vector spanned by the ${u,v,w}$.

So I can see that the vector sum u+v+w cannot be linearly dependent because of the fact that any vector can be written as a sum of these three vectors and at least one non-zero coefficient from {u,v,w} being a basis. It also follows in the same manner the v+w and w are not linear dependent sums. The problem I have is showing that the sum of all three of these is linearly independent. Hmmm...

As for spanning vector space V, let's say that $a_1u+a_2w+a_3v$ can represent any vector in V by definition. Then $a_1u$ can be written in terms of the "basis" we are trying to prove as $a_1(u+v+w)-a_1(v+w)$, $a_2w$ as simply $a_2w$, and $a_3v$ as $a_3(v+w)-a_3w$.

Look ok minus the missing part which I need help on?

2. Originally Posted by Jameson
Let ${u,w,v}$ be a basis for vector space V. Prove that ${u+v+w, v+w,w}$ is also a basis for vector space V
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Ok so the show that this is a basis, it must satisfy being linearly independent and must be able to span any vector spanned by the ${u,v,w}$.

So I can see that the vector sum u+v+w cannot be linearly dependent because of the fact that any vector can be written as a sum of these three vectors and at least one non-zero coefficient from {u,v,w} being a basis. It also follows in the same manner the v+w and w are not linear dependent sums. The problem I have is showing that the sum of all three of these is linearly independent. Hmmm...

As for spanning vector space V, let's say that $a_1u+a_2w+a_3v$ can represent any vector in V by definition. Then $a_1u$ can be written in terms of the "basis" we are trying to prove as $a_1(u+v+w)-a_1(v+w)$, $a_2w$ as simply $a_2w$, and $a_3v$ as $a_3(v+w)-a_3w$.

Look ok minus the missing part which I need help on?
Your argument for spanning looks solid to me.

To prove independence, I think this is what you want to do:

Independent:

Show that the only solution to $\alpha (u + v + w) + \beta (v + w) + \gamma w = 0$ is $\alpha = \beta = \gamma = 0$.

$\alpha (u + v + w) + \beta (v + w) + \gamma w = 0$

$\Rightarrow \alpha u + (\alpha + \beta) v + (\alpha + \beta + \gamma ) w = 0$.

Since u, v and w are a basis it follows that

$\alpha = 0$ ... (1)

$\alpha + \beta = 0$ ... (2)

$\alpha + \beta + \gamma = 0$ ... (3)

The solution to equations (1), (2) and (3) is clearly $\alpha = \beta = \gamma = 0$ and so independence is proved.