# Prove an orthogonal dual basis without using coordinates

Let $$V$$ be a vector space over a base field $$\mathbb F$$. Let $$V'$$ be the dual vector space of all linear functionals $$V \to \mathbb F$$. If $$v_1,\ldots, v_n$$ form a basis for $$V$$, I claim there is a dual basis for $$V'$$ given by $$v_1',\ldots,v_n'$$ for which $$v_i'\left(v_j\right) = \delta_{ij}$$.

This all feels a bit circular.

Let $$e_1,\ldots,e_n$$ be the standard basis on $$V$$ and $$e_1',\ldots,e_n'$$ the standard basis for $$V'$$, where $$e_i'(e_j)=\delta_{ij}$$. See how circular this is. I've assumed what I want to show. In coordinates, with $$n=3$$ then $$e_1=(1,0,0)^{\top}$$, $$e_2 = (0,1,0)^{\top}$$ and $$e_3=(0,0,1)^{\top}$$; general vectors are $$x~e_1+y~e_2+z~e_3$$, where $$x,y,z \in \mathbb F$$. In the case of the dual: $$e_1' = \mathrm dx$$, $$e_2' = \mathrm dy$$ and $$e_3' = \mathrm dz$$; general covectors are $$p~\mathrm dx + q~\mathrm dy+r~\mathrm dz$$, where $$p,q,r \in \mathbb F$$.

Now I need to use coordinates:

Let a basis for $$V$$ be $$v_1=e_1+e_2-e_3$$, $$v_2=e_1-e_2+e_3$$ and $$v_3=-e_1+e_2+e_3$$. I seek covectors $$v_1',v_2'$$ and $$v_3'$$ for which $$v_i'\left(v_j\right) = \delta_{ij}$$. Identifying vectors with $$3\times 1$$ column vectors and covectors with $$1\times 3$$ row vectors gives

$$\left(\begin{array}{ccc} \leftarrow & v_1' & \rightarrow \\ \leftarrow & v_2' & \rightarrow \\ \leftarrow & v_3' & \rightarrow \end{array}\right)\left(\begin{array}{ccc} 1 & 1 & -1 \\ 1 & -1 & 1 \\ -1 & 1 & 1\end{array}\right) = I_3$$ where $$I_3$$ is the $$3 \times 3$$ identity matrix.

Hence:

$$\left(\begin{array}{ccc} \leftarrow & v_1' & \rightarrow \\ \leftarrow & v_2' & \rightarrow \\ \leftarrow & v_3' & \rightarrow \end{array}\right) = \left(\begin{array}{ccc} 1 & 1 & -1 \\ 1 & -1 & 1 \\ -1 & 1 & 1\end{array}\right)^{-1}$$

$$\left(\begin{array}{ccc} \leftarrow & v_1' & \rightarrow \\ \leftarrow & v_2' & \rightarrow \\ \leftarrow & v_3' & \rightarrow \end{array}\right) = \left(\begin{array}{ccc} 1/2 & 1/2 & 0 \\ 1/2 & 0 & 1/2 \\ 0 & 1/2 & 1/2 \end{array}\right)$$

$$v_1' = \frac{1}{2}~\mathrm dx+\frac{1}{2}~\mathrm dy$$, $$v_2'=\frac{1}{2}~\mathrm dx+\frac{1}{2}~\mathrm dz$$ and $$v_3'=\frac{1}{2}~\mathrm dy+\frac{1}{2}~\mathrm dz$$.

I now have a dual basis where $$v'_i\left(v_j\right) = \delta_{ij}$$.

How do I prove such a basis exists without resorting to coordinates, identifying covectors with rows and vectors with columns, and without inverting coefficient matrices? Is it not circular to assume a basis $$e_1,\ldots,e_n$$ and $$e_1',\ldots,e_n'$$ with $$e_i'\left(e_j\right)=\delta_{ij}$$?

• Your definition of the functionals $e_i'$ is not circular: it is a fact that any function defined over a basis can be uniquely extended to a linear map. We are defining the $i$th map $e_i'$ by what it does to the basis $e_1,\dots,e_n$. Commented Dec 11, 2023 at 20:33
• I'd say that, having established what these linear maps are, what remains to be seen is that they form a basis of $V'$. Commented Dec 11, 2023 at 20:34
• @FlybyNight Perhaps you misinterpreted my first two comments: my intent was to outline the start of a proof. As I said, you can begin with your definitions of the elements $e_1',\dots,e_n' \in V'$; contrary to what you say, your definition of these functionals is not circular. Having established what these are, you should then show that they are linearly independent and span $V'$. Commented Dec 11, 2023 at 20:53
• @FlybyNight Yes. With that, you have shown that these elements are linearly independent. From there, you could either use the fact that $\dim(V) = \dim(V')$ or directly show that the elements $v_1',\dots,v_n'$ span $V'$. Commented Dec 11, 2023 at 21:00
• @FlybyNight I'll write something up Commented Dec 11, 2023 at 21:04

We begin with a basis $$v_1,\dots,v_n$$ of $$V$$. We then define a set of linear maps $$v_i':V \to \Bbb F$$ by $$v_i'(v_j) = \delta_{ij}$$. Contrary to what you say in the question, there is nothing "circular" about defining linear maps in this way: this is an application of the fact that there is a unique linear extension to any function over a basis (cf. this post for instance). However, there is still the matter of showing that the functionals defined in this way form a basis, i.e. that they are linearly independent and span $$V$$.
To show that $$(v_1',\dots,v_n')$$ is a linearly independent sequence, consider any linear combination $$f = \lambda_1 v_1' + \cdots + \lambda_n v_n' = 0$$. We see that for any $$i$$, $$0 = f(v_i) = \lambda_i.$$ Thus, $$\lambda_1 v_1' + \cdots + \lambda_n v_n' = 0$$ implies that $$\lambda_1 = \cdots = \lambda_n = 0$$.
To show that $$(v_1',\dots,v_n')$$ is a spanning set, consider any $$f \in V$$. I claim that the linear combination $$g = f(v_1)v_1' + \cdots + f(v_n)v_n'$$ is equal to $$f$$. To see that this is the case, show that $$f(v_i) = g(v_i)$$ for $$i = 1,\dots,n$$, then once again use the fact that linear functions are uniquely determined by what they do to the elements of a basis.