# Does my proof show a further result?

NOTE: My actual question is in the last paragraph. Feel free to skip down there first to see what exactly I'm asking.

Here's the question I've written a proof for:

Suppose $V$ is finite-dimensional and $\varphi_1, \dots, \varphi_n$ is a basis of $V'$. Show that there exists a basis of $V$ whose dual basis is $\varphi_1, \dots, \varphi_n$.

I have written a proof that I think is correct. Here it is:

By $\widehat{\operatorname{null}\varphi_j}$ denote $\operatorname{null}\varphi_1 \cap \cdots \cap \operatorname{null}\varphi_{j-1}\cap\operatorname{null}\varphi_{j+1}\cap \cdots \cap \operatorname{null}\varphi_n$. From [the previous exercise], we see that $\dim\widehat{\operatorname{null}\varphi_j} = 1$ for all $j=1,\dots, n$. I.e. for any $j=1,\dots, n$, there exists a nonzero $v_j\in\widehat{\operatorname{null}\varphi_j}$. As a further consequence of [the previous exercise], we see that $v_j\not\in \operatorname{null}\varphi_j$ because $\dim(\operatorname{null}\varphi_1\cap \cdots \cap \operatorname{null}\varphi_n) = \dim V - \dim V' = 0$ (i.e. $\operatorname{null}\varphi_1\cap \cdots \cap \operatorname{null}\varphi_n = \{0\}$). It follows that $\left\{\frac{v_1}{\varphi_1(v_1)}, \dots, \frac{v_n}{\varphi_n(v_n)}\right\}$ has the property $\varphi_j\left(\frac{v_i}{\varphi_i(v_i)}\right) = \delta_{ij}$ as desired.

To show that this is a basis it suffices to show that it is linearly independent. Let $a_1, \dots, a_n\in \Bbb F$ such that $$a_1\frac{v_1}{\varphi_1(v_1)} + \cdots + a_n\frac{v_n}{\varphi_n(v_n)} = 0.$$ Applying $\varphi_j$ to both sides yields $a_j=0$ for any $j=1,\dots, n$, completing the proof.$\square$

Now here's my question. This exercise is clearly stated as an existence proof. With my particular proof, have I also shown uniqueness? Afterall, while the choice of $v_j$ was arbitrary, it was in a $1$-dimensional space $\widehat{\operatorname{null}\varphi_j}$, which is just the multiples of this vector. Then I (kinda) normalized it by dividing by $\varphi_j(v_j)$. Meaning if instead I'd chosen $kv_j$, I'd just have $\frac{kv_j}{\varphi_j(kv_j)} = \frac{kv_j}{k\varphi_j(v_j)} = \frac{v_j}{\varphi_j(v_j)}$ in my basis. So doesn't that mean that each $\frac{v_j}{\varphi_j(v_j)}$ is unique?

Yes, the basis $\{w_1,\dots,w_n\}$ of which $\{\varphi_1,\dots,\varphi_n\}$ is the dual basis is uniquely determined. Indeed, if $v$ is a vector such that $\varphi_i(v)=1$ and $\varphi_j(v)=0$ for $j\ne i$, then $$\varphi_j(v-w_i)=0$$ for every $j=1,2,\dots,n$. In particular, every linear form vanishes over $v-w_i$, so $v-w_i=0$. A property of the dual space is that, for every nonzero vector, there is a linear form that doesn't vanish on it.
There is a possibly better proof. Consider the dual basis in $V''$ and pull it back to $V$ via the canonical isomorphism $V\to V''$.
• That's pretty clever (the top part, I don't get to the canonical isomorphism between $V$ and $V''$ for two more exercises). Accepted. – Bobbie D Mar 4 '17 at 21:35