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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 $e_i'\left(e_j\right)=\delta_{ij}$?

Fly by Night
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  • 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 $e_1,\dots,e_n$. – Ben Grossmann Dec 11 '23 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 $V'$. – Ben Grossmann Dec 11 '23 at 20:34
  • @BenGrossmann I've added some matrices. If $M$ is the matrix whose columns are the coefficients of the $e_i$, e.g. $v_1=2e_1+3e_2+4e_3$ gives a column of $(2,3,4)^{\top}$. The coefficients in the expression of the $v_i'$ are given by the rows of $M^{-1}$, see above. If $\det M \neq 0$, then $\det\left(M^{-1}\right) \neq 0$, so the $v_1',\ldots,v_n'$ span $V'$. – Fly by Night Dec 11 '23 at 20:40
  • I don't understand what the specific matrix is supposed to illustrate or change about the question being asked. Do my comments not address the question you have in mind? Is there something I said that requires elaboration? – Ben Grossmann Dec 11 '23 at 20:42
  • I thought you were asking me to show that the $v_1$, $v_2$ and $v_3$ I constructed above span $V'$, and they do because their coefficient matrix is non-singular. I'd really like to be able to prove that for a basis $v_1,\ldots,v_n$ of $V$, there is a basis of $V'$, say $v_1',\ldots,v_n'$ with $v_i'(v_j) = \delta_{ij}$, but without resorting to coefficient matrices. – Fly by Night Dec 11 '23 at 20:48
  • You do not need the standard basis (or its coordinates) to derive the dual basis to a different basis. It is an abstract construction. If you want to identify the "new" basis by its coordinates in the standard basis, then of course you're going to use coordinates to find the coordinates of the dual basis. – Ted Shifrin Dec 11 '23 at 20:49
  • @TedShifrin How do I prove the following statement: "Given a basis ${v_1,\ldots,v_n}$ of an $\mathbb F$-vector space $V$, there exists a basis ${v_1',\ldots,v_n'}$ of the dual vector space $V'$ of linear functionals $V \to \mathbb F$, for which $v_i'(v_j) = \delta_{ij}$"? – Fly by Night Dec 11 '23 at 20:52
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    @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 $V'$. – Ben Grossmann Dec 11 '23 at 20:53
  • @BenGrossmann So, assume there are linear functionals $v_1', \ldots,v_n'$ with the property that $v_i'(v_j)=\delta_{ij}$ and $\lambda_1v_1'+\cdots+\lambda_nv_n'=0$ for $\lambda_i \in \mathbb F$. Now I want to show $\lambda_1=0, \ldots, \lambda_n = 0$? Applying this to $v_i$ in turn gives $(\lambda_1v_1'+\cdots+\lambda_nv_n')(v_i) = 0$ and so $\lambda_i = 0$. – Fly by Night Dec 11 '23 at 20:57
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    @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 $V'$. – Ben Grossmann Dec 11 '23 at 21:00
  • Consider two column vectors $x$ and $y$ as matrices. If we consider the equation $x^Ty$ with the usual matrix multiplication this is the dot product. If you hold $x$ constant and let $y$ vary then you get a vector space. Holding $y$ constant and allowing $x$ to vary gives you its dual. So if it looks a little circular it's because in finite dimensions the distinction is mostly notational. – CyclotomicField Dec 11 '23 at 21:01
  • @BenGrossmann Do you want to write that up as an answer and I'll accept it, or should I write it up and mention you at the start? – Fly by Night Dec 11 '23 at 21:04
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    @FlybyNight I'll write something up – Ben Grossmann Dec 11 '23 at 21:04

1 Answers1

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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.

Ben Grossmann
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