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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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    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

3 Answers 3

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In general I find it much harder to show that a set of vectors spans the vector space, than showing a set is independent. If you want to go about your approach and show they span, what you would need to do is take an arbitrary vector in $V$ and write it is a linear combination of your new set.

An alternative way is to show that they are independent, which turns out to be quite simple.

Suppose $c_1(u + v + w) + c_2(v+w) + c_3(w) = 0$ then we know that $c_1u + (c_1 + c_2)v + (c_1 + c_2 + c_3) w = 0$ but $u, v, w$ are independent and so $c_1 = c_1 + c_2 = c_1 + c_2 + c_3 = 0$ From here its pretty clear that $c_1 = c_2 = c_3 = 0$ which would prove the claim.

EDIT: Note, if we show that the vectors are independent they must also span $V$, because in any vector space of dimension $n$, if we have $n$ independent vectors, they must span the vector space. Similarly, if we were able to show that they spanned the space then they would have to be independent since we have $\dim(V)$ of them

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    Nice alternative2011-12-06
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It is easily seen that each of your original basis elements are in the span of $\{u+v+w,v+w,w\}$. Since $\{u, v, w\}$ spans $V$, so does $\{u+v+w,v+w,w\}$ (if $A\subseteq {\rm span} \,(V)$, then ${\rm span}(A)\subseteq{\rm span }(V) \thinspace $).

So, we have a set of three vectors that span a space of dimension three. It follows that the set is independent and thus a basis.

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The proof is essentially correct, but you do have some unnecessary details. Removing redundant information, we can reduce it to the following:

Let $V$ be a vector space over $F$ and $\{u, v, w\}$ a basis for $V$. Any $x \in V$ can be uniquely written $x = au + bv + cw$ for some $a, b, c \in F$. Let $d, e, f \in F$ be the unique choices so that $d = a$, $e = b - a$, and $f = c - b$; then $x$ can be uniquely written as $x = du + (d + e)v + (d + e + f)w$. Rearranging, we have $x = d(u + v + w) + e(v + w) + fw$. Since this is a unique expression for an arbitrary $x \in V$, then $\{u + v + w, v + w, w\}$ is a basis for $V$.

There is also this nice alternate proof:

Let $V$ be a vector space over $F$ and $\{u, v, w\}$ a basis for $V$. Let's start a new basis with $w$. Since $\{u, v, w\}$ is linearly independent, $v + w \notin span\{w\}$ and $u + v + w \notin span\{w, v + w\}$, so $\{w, v + w, u + v + w\}$ is linearly independent. It is also the same length as a basis, so it must be a basis itself.

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    since $x = au + bv + cw$ is unique and each $(a, b, c)$ corresponds to exactly one $(d, e, f)$, then $x = d(u + v + w) + e(v + w) + fw$ must also be unique.2011-12-06