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Let $V$ be a finite $n$-dimensional vector space over a field $K$ and $\{\lambda_{1},\ldots, \lambda_{n}\}$ be a linearly independent set of functionals

Show that the linear map

$$\Lambda:V\to K^n$$ $$v\mapsto (\lambda_1(v),\ldots, \lambda_n(v))$$

Is surjective.

This is a detail in a proof of a proposition related to dual bases which I simply can't solve.

If we let $\{\lambda_1,\ldots, \lambda_n\}$ be a linearly independent then we have that if we remove one functional from that set the span of the rest of the vectors does not contain the removed element, or equivalently if $a_1\lambda_1 + \cdots + a_1\lambda_n = 0$ where not all functionals are nonzero, $a_j \in K$ and $0$ is the zero functional taking all elements of $V$ to $0\in K$ then $a_1 = \cdots = a_n = 0$

If $\Lambda$ is surjective then the dimension of the image of $\Lambda$ but be equal to $n$ which implies that the kernel of the map is trivial so my failed strategy was to show that that the kernel is trivial from which I would conclude that the map must be trivial.

If $v\in \ker(\Lambda) \implies av \in \ker(\Lambda) \forall a \in K$ since the the kernel is a subspace (closure).

$$\Lambda(av) = (0,\ldots, 0) \implies a(\lambda_1(v),\ldots, \lambda_n(v)) = (0,\ldots, 0)$$

So $a(\lambda_1(v) + \cdots + a\lambda_n(v)) = 0$ But I don't get anyway from here and realize that I've probably done some misstake or thinking about the porblem in a stupid way.

Any hints would be appreciated.

Michael Hardy
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John Smith
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    If you write \lambda\text{ker}\Lambda you see $\lambda\text{ker}\Lambda$ but if you write \lambda\ker\Lambda you see $\lambda\ker\Lambda$, with proper spacing before and after "ker". Proper spacing includes the fact that in $\lambda\ker(\Lambda)$ there shouldn't be as much space after "ker" as if the left parenthesis were not there. Using \ker in that way is standard usage. I changed it to \ker. $\qquad$ – Michael Hardy Feb 10 '16 at 16:30

1 Answers1

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Suppose $(\lambda_1(v),\ldots, \lambda_n(v)) = 0\in V$. Then $\lambda_1(v)=\cdots=\lambda_n(v)=0$. Therefore every linear combination $a_1\lambda_1(v)+\cdots+a_n\lambda_n(v)$ is $0\in V$. If $v$ is in the kernel of every linear combination of $\lambda_1,\ldots,\lambda_n$ then $v$ is in the kernel of every linear functional on $V$, since $\{\lambda_1,\ldots,\lambda_n\}$ spans the whole space of linear functionals on $V$. (Here I'm assuming you've seen it proved somewhere that the dimension of the space of linear functionals on $V$ is the same as the dimension of $V$.) The only vector in $V$ that is in the kernel of every linear functional on $V$ is $0$. (In other words if $0\ne v\in V$ then some linear functional maps $v$ to some nonzero scalar.)

Therefore, if $(\lambda_1(v),\ldots, \lambda_n(v)) = 0\in V$, then $v=0$.

So the kernel of $v\mapsto(\lambda_1(v),\ldots, \lambda_n(v))$ is trivial.

Michael Hardy
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