Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $v_1, \ldots, v_n$ be a set of vectors in a vector space $V$. Show that $v_1, \ldots, v_n$ is a basis of $V$ if and only if for any non-zero linear function $f$ on $V$ there is a vector $v$ in $\operatorname{span}(v_1, \ldots, v_n)$ such that $f(v) \neq 0$.

Suppose that $v_1, \ldots, v_n$ is not a basis of $V$. Then the complement $W$ of $\operatorname{span}(v_1, \ldots, v_n)$ in $V$ is not empty. Let $f$ be a function such that $f(x)=1$ for all $x\in W$ and $f(x)=0$ for all $x$ in $\operatorname{span}(v_1, \ldots, v_n)$. My question is: is $f$ linear in $V$?

share|improve this question
    
No, $f$ is not linear unless $W$ is empty: pick $w\in W$, and let $x=w$; then $f(2x) = 1$ (because $2x\in W$), but $2f(x) = 2$, so $f$ is not homogeneous. It is also not additive: take $x=w$, and $y=v_1-w$. Then $f(x)=f(y)=1$, but $f(x+y) = f(v_1) = 0\neq f(x)+f(y)$. –  Arturo Magidin Dec 13 '11 at 21:52

3 Answers 3

up vote 1 down vote accepted

Every linear transformation is determined by its values on a basis. These values may be chosen arbitrarily. So, your $f$ is a linear functional on $V$ if you require that $f(x)=1$ for a basis of $W$, not for the whole of $W$. A constant function cannot be linear unless it is zero.

share|improve this answer
    
$W$ is not a subspace: it's the complement of a subspace, so it doesn't even have a zero vector. What is "a basis for $W$"? –  Arturo Magidin Dec 13 '11 at 21:57
    
Oh, I read $W$ as a linear complement of $\operatorname{span}(v_1, \ldots, v_n)$, ie, a subspace such that $V=W \bigoplus\operatorname{span}(v_1, \ldots, v_n)$. –  lhf Dec 13 '11 at 21:59
    
I guess that's a possible reading... fair enough –  Arturo Magidin Dec 13 '11 at 22:00

Um, unless I'm misreading the question, the thing to be proved is false. Counterexample: Let $V=\mathbb R^1$ (a real vector space) and let $n=2$, $v_1=5$, and $v_2=42$. Then $\mathrm{span}(5,42)$ is $\mathbb R^1$ itself, and it is certainly the case that for any nonzero linear $f$, there is a vector $v$ in $\mathbb R^1$ such that $f(v)\ne 0$ -- that is the definition of $f$ being nonzero.

But the set $(v_1,v_2)$ is not a basis for $\mathbb R^1$ because it is linearly dependent; to wit, $42v_1-5v_2=0$.

share|improve this answer

As Henning points out, the statement is false as written; but if you add the assumption that $v_1,\ldots,v_n$ are linearly independent, then the result is true.

As to your idea, it's either completely wrong or almost right, depending on what you mean by "complement."

If by "complement" you mean the set-theoretic complement (that is, $W$ is the set of all vectors in $V$ that are not in $\mathrm{span}(v_1,\ldots,v_n)$), which was my interpretation when I read your post, then your approach is completely wrong: this set is not closed under sums, and in general you cannot expect it to be well-behaved.

If by "complement" you mean a linear complement (as lhs understood your writing), that is, a subspace $W$ such that $W\cap\mathrm{span}(v_1,\ldots,v_n)=\{\mathbf{0}\}$ and $W+\mathrm{span}(v_1,\ldots,v_n) = V$), then you are almost right: you cannot define $f$ as $1$ on all of $W$ (that would not be homogeneous or additive); but you can find a basis for $W$ (say, by extending $v_1,\ldots,v_n$ to a basis of $V$, $v_1,\ldots,v_n,v_{n+1},\ldots,v_{n+m}$ and letting $W=\mathrm{span}(v_{n+1},\ldots,v_{n+m})$), define $f$ as $1$ on that basis, and then "extending linearly".

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.