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Let $u,v,w \in V$ a vector space over a field F such that $u \neq v \neq w$. If $\{ u , v , w \}$ is a basis for $V$, then prove that $\{ u+v+w , v+w , w \}$ is also a basis for $V$.

Proof:

Let $u,v,w \in V$ a vector space over a field $F$ such that $u \neq v \neq w$. Let $\{ u , v , w \}$ be a basis for $V$. Because $\{ u , v , w \}$ is a basis, then $u,v,w$ are linearly independent and $ \langle \{ u , v , w \} \rangle = V$.

Let $x \in V$ be an arbitrary vector then $x$ can be uniquely expressed as a linear combination of $\{ u , v , w \}$. Let's suppose $x=au+bv+cw$ for some $a,b,c \in F$.

On the other hand, let us consider $\{ u+v+w , v+w , w \} \subseteq V$.

Then $$ \begin{align*} \langle \{ u+v+w , v+w , w \} \rangle &= \{d(u+v+w) + e(v+w) + f(w) \mid d,e,f \in F\} \\ &= \{du + (d+e)v +(d+e+f)w \mid d,e,f \in F \} . \end{align*}$$

If $x \in V$, then $x=du + (d+e)v +(d+e+f)w$ is another unique representation of $x \in V$ . Then for any arbitrary $x \in V$, we have $d=a$, $d+e=b $and $d+e+f=c \in F$.

Because $\{ u , v , w \}$ is a basis for $V$, then $\{ u+v+w , v+w , w \}$ must also be a basis for $V$.

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    I tried to give an alternate proof instead of proving <{ u , v , w }>=V<{ u+v+w , v+w , w }>2011-12-06
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    Note that $u\neq v\neq w$ only means that $u\neq v$ and $v\neq w$, and does not mean that also $u\neq w$ ("is not equal to" is not transitive). Note also that saying that the vectors are unequal is redundant when you assume they form a basis. No two elements of a linearly independent set can be equal.2011-12-06
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    Thanks. I typed that because that is how the statement was given. I do undestand that no two elements of a linearly independent set can be equal.2011-12-06
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    Are you aware of the fact that you can type LaTeX commands for your math notation?2011-12-06
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    I should say that you are very, very close to proving that they do span. You just need to take in an arbitrary triple $a, b, c$ and produce $d, e, f$ such that the equations in your penultimate paragraph hold.2011-12-06
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    Isn't saying that for for any arbitrary x∈V=<{ u , v , w }> with d=a, d+e=b and d+e+f=c it implies the unique solution d=a, e=b-d=b-a , f=c-d-e=c-a-(b-a)=c-a-b+a=c-b ? My idea was exactly to take something in the span of {u+v+w,v+w,w} and writte it as a linear combination of my original basis elements with the above unique expresion for any x∈V2011-12-06
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    @AndresOrtizMena you say "My idea was exactly to take something in the span of $\{u + v + w, v + w, w\}$ and write it as a linear combination of my original basis elements" but you can always do this, simply because the original basis, is a basis. What you need to do is take something arbitrary in the span of your original basis (namely in $V$) and write it as a linear combination of your new set.2011-12-06
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    @Andres I think we're just misinterpreting each other. I agree with your first sentence and I think you have a fine strategy; I just couldn't tell how you were concluding. I would add the assignment of $d, e, f$ to your proof. You should probably spell out why $d, e, f$ are uniquely determined by $x$, as well.2011-12-06
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    because for any arbitrary x in V we have a unique representation of x with the original basis and we can make another unique representation of x as a linear combination of my new basis thus we can find those unique constants looking that the span of my new basis can be "stated" as a linear combination of the old one so the constants can be uniquely determined2011-12-06
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    Another proof, essentially equivalent to the others suggested, but using different technology, would be to write the relationship between the original base and the new combination in matrix form, and then write down the inverse. In more complex cases a non-zero determinant would mean an invertible transformation, and the rank of the matrix would give the dimension of the image. But I guess this technology comes later in the development of the subject.2011-12-06

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