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We have a $\mathbb{K}$-vector space $V$, and $\mathcal{F}_{\mathbb{K}}(V,\mathbb{K})$ is the $\mathbb{K}$-vector space of all functions $f:V \to \mathbb{K}$, with the operations $(f_{1} + f_{2})(v) = f_{1}(v) + f_{2}(v)$ and $\alpha f(v) = (\alpha f)(v)$.

$V^* = Hom_{\mathbb{K}}(V, \mathbb{K}) \subset \mathcal{F}_{\mathbb{K}}(V,\mathbb{K})$ is the subspace that contains all homomorphisms $T:V \to \mathbb{K}$.

$B = \{v_{1},...,v_{n}\}$ is a basis of $V$. If $u \in V$ and $u = \sum_{i=1}^n x_{i}v_{i}$, we define the function $\phi_{j}\in V^*$ such that $\phi_{j}(u) = x_{j}$.

I'm asked to prove that $B^* = \{\phi_{1},...,\phi_{n}\}$ is a basis of $V^*$, but I'm having trouble showing that $B^*$ is linearly independent. I tried making a sum:

$$(\sum_{i = 1}^n \beta_{i}\phi_{i})(u) = \sum_{i = 1}^n (\beta_{i}\phi_{i})(u) = \sum_{i = 1}^n \beta_{i}(\phi_{i}(u)) = \sum_{i = 1}^n \beta_{i}x_{i} = \mathbb{O}(u) = 0$$

I don't see how a linear combination of the coordinates of a vector necessarily implies that all $\beta_{i}$ must be zero.

Juanma Eloy
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3 Answers3

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Note that $\forall 1\le i,j\le n:\, \phi_j(v_i)=\begin{cases} 0\iff i\ne j\\ 1\iff i=j\end{cases}$

So considering the linear combination of $\phi_j$’s $f=\sum_{j=1}^n a_j\phi_j$ you see that $\forall 1\le i\le n:\, f(v_i)=a_i$. Thus to obtain $f\equiv 0$ that must be $\forall 1\le i\le n\,\, a_i=0$.

Alessandro Cigna
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You just need to prove that $B^*$ spans $\textsf V^*$, that is, for any $f\in \textsf V^*$ we have that $$f=\sum_{i=1}^n a_i \phi_i$$ for some scalars $a_1,a_2,\dots,a_n$ (in fact, $a_i=f(v_i)$). The linear independence follows that $\dim(\textsf V^*)=n$.

You can also check mi comment right here.

azif00
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Note that for all $i,j \in \{1, \dots, n\}$ we have the nice property $\phi_i(v_j) = \delta_{ij}$, where $\delta_{ij}$ is the kronecker-delta symbol which is $1$ if $i=j$ and $0$ if $i\neq j$.

Let $a_1, \dots, a_n \in \Bbb{K}$ be scalars such that \begin{align} \sum_{i=1}^na_i \phi_i = 0. \end{align}

Now, to prove linear independence, we need to show that for each index $j \in \{1, \dots, n\}, $ we have $a_j = 0$. To prove this, we shall apply the above equation on the vector $v_j$. Doing this gives us \begin{align} 0 &= \sum_{i=1}^n a_i \phi_i(v_j) \\ &= \sum_{i=1}^n a_i \delta_{ij} \\ &= a_j \end{align} Since this is true for all $j \in \{1, \dots, n\}$, it follows that $\{\phi_1, \dots, \phi_n\}$ is linearly independent.

You said

I don't see how a linear combination of the coordinates of a vector necessarily implies that all $\beta_i$ must be zero.

This is true, you cannot conclude this directly. But the problem with your approach is you applied the function on an arbitrary vector $u$. What you should have done is chosen specific vectors (the $v_j$'s) and use the defining property of the $\phi_i$'s to conclude that each coefficient is in fact zero.

peek-a-boo
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