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Prove that for any non-zero linear function $f$ of a $n$-dimensional vector space $V$ there exists a basis ${e_1, ..., e_n}$ for the $V$ vector space, if

$$f(x_1e_1+ ... + x_ne_n) = x_1$$ should be true for any $x_1, ..., x_n$

How to prove that? Thanks.

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What are you trying to prove? What has $f$ go to do with it? Do you mean 'show that there exists a basis $e_k$ such that $f(e_1) = 1$'? –  copper.hat Feb 23 '13 at 19:17
    
@copper.hat This is equivalent to the existence of a basis with $f(e_1)=1$, and $e_2,,\ldots,e_n$ in the nullspace of $f$. –  1015 Feb 23 '13 at 19:43
    
@julien: I realize that, I was trying to clarify the wording. –  copper.hat Feb 23 '13 at 19:49
    
@copper.hat I agree, this "if ...should be true" could be ameliorated. A good old "such that" would probably be better. But the question is fairly clear. And in particular what $f$ has to do with it, no? –  1015 Feb 23 '13 at 19:59
    
@julien: It seemed like a possible question, but wasn't clear to me. That's why I added the first comment. I have spend hours sometimes answering what seemed like a fairly straightforward question, only to have a (linguistically) minor tweak render the hours pointless. Now my default reaction is to clarify first. –  copper.hat Feb 23 '13 at 20:03
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2 Answers

up vote 1 down vote accepted

So $V$ is an $n$-dimensional $K$ vector space. This fact you want to prove is true for any field $K$.

Since $f$ is nonzero, its range must be $K$. Therefore the nullspace of $f$ has dimension $n-1$ by the rank-nullity theorem.

Let $e_2,\ldots,e_n$ be a basis of this nullspace.

Then pick any $e_1$ such that $f(e_1)=1$.

It follows that $e_1,e_2,\ldots,e_n$ is a basis of $V$.

Indeed, if you take a linear combination of the latter and apply $f$, you'll see that the coefficient of $e_1$ is equal to $0$. Then you can use the linear independence of $e_2,\ldots,e_n$ to get the nullity of the other coefficients. So $e_1,\ldots, e_n$ are linearly independent. Given the dimension, they must form a basis of $V$.

Now $$ f(x_1e_1+\ldots+x_ne_n)=\sum_{k=1}^n x_kf(e_k)=x_1 $$ as you wanted.

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Here is a more tedious approach that is slightly constructive.

Let $v_k$ be a basis for $V$. Choose $w_1 \in V$ such that $f(w_1) = 1$ ($f$ is non-zero, and linear, so we can do this, and $w_1$ is necessarily non-zero). Let $w_1 = \sum \alpha_k v_k$, and suppose for convenience that $\alpha_1 \neq 0$ (at least one of the $\alpha_k$ is non zero, so we can always do this).

Then it is straightforward to show that $w_1, v_2,...,v_n$ is a basis for $V$.

Now let $w_k = v_k -f(v_k) w_1$, for $k >1$, and note that $f(w_k) = 0$.

Then $w_1,...,w_n$ is a basis for $V$ with the required property. To check that the $w_k$ are linearly independent we note that $\sum_k \alpha_k w_k = (\alpha_1-\sum_{k>1}\alpha_kf(v_k) )w_1 + \sum_{k>1} \alpha_k v_k$, hence if $\sum_k \alpha_k w_k = 0$, then $\alpha_k = 0$.

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