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Linear algebra in elementary calculus

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Just an example...

  • Given a continuous function $f \colon {\bf R} \rightarrow {\bf R}$,
  • Let $\Phi$ be the set of all closed intervals in ${\bf R}$.
  • Define $F \colon \Phi \rightarrow {\bf R}$ by

$$F([a,b]) = \displaystyle\int_a^b f(x)dx.$$ Now, in what sense is it linear?

In other words, how can we make it linear?

Define: $$\Phi^* = {\rm span \hspace{3pt}} \Phi,$$ in the sense that it is the set of all "formal" linear combinations of $\Phi$, which serves as a basis. The obvious choice of operations makes $\Phi^*$ a vector space.

Next, we define $$F^* \colon \Phi^* \rightarrow {\bf R}$$ by $$F^*(\displaystyle\sum_{i=1}\alpha_i [a_i,b_i]) = \displaystyle\sum_{i} \alpha_i F([a_i,b_i]).$$ A function defined this way, in terms of a given basis, is "automatically" linear.

Now we incorporate additivity and orientation of integral: $$\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx,$$ $$\int_a^b f(x)dx = -\int_b^a f(x) dx.$$

We define an equivalence relation based on these properties:

  • Additivity: $[a,b] + [b,c] \sim [a,c]$.
  • Orientation: $[a,b] \sim -[b,a]$.

We want to show $\Phi^*/_{\sim}$ is a vector space too.

What's required?

This new vector space can more commonly defined via a certain subspace. Indeed, consider: $$\Phi^*/_{\sim} = \Phi^*/V,$$ where $V$ is a subspace with $V \sim 0$.

We need to define $V$. How?

Consider the additivity property first: $$[a,b] + [b,c] \sim [a,c],$$ or $$[a,b] + [b,c] - [ac] \sim 0.$$ This is the first step, $V_1$, toward $V$.

So let $$V_1 = {\rm span} \{ [a,b] + [b,c] - [a,c] \colon a,b,c \in {\bf R}\} \subset \Phi^*.$$

Is this a subspace?

It's closed under operations (two of them...), so yes, by a theorem from linear algebra.

Next we consider the orientation property: $$[a,b] \sim -[b,a],$$ or $$[a,b] + [b,a] \sim 0.$$ This is the second step, $V_2$.

So let $$V_2 = {\rm span} \{ [a,b] + [b,a] \colon a,b \in {\bf R}\}.$$ Prove that $V_2$ is a subspace. Then let $$V = {\rm span} (V_1 \cup V_2),$$ and $\Phi^{**} = \Phi^*/V$.

Question: Is $F \colon \Phi^{**} \rightarrow {\bf R}$ well defined?

Note: we need a new $F^{**} \colon \Phi^{**} \rightarrow {\bf R}$, because $F^* \colon \Phi^* \rightarrow {\bf R}$ and $F^{**}$ have different domains.

If $a \sim b$ then $F^*(a) = F^*(b)$, then $F^{**}([a]) = F^*(a) = F^*(b)$, so yes!


  • the integral as a function $\displaystyle\int \hspace{3pt} \colon {\rm closed \hspace{3pt} intervals} \rightarrow {\bf R}$ is linear.

Now the punch line...

Question: What is $F^* \colon \Phi^* \rightarrow {\bf R}$?

Answer: It's a $1$-form!

Indeed, it's a linear function on intervals ($1$-cells) with anti-symmetry.

This is an alternative way to introduce forms -- via integration. No formulas, just each input given you an output. Like this:


You can think of this as algebraic representation of functions vs. numerical representation of functions.

This fits into a bigger story:

  • the integral as a function $\displaystyle\int \hspace{3pt} \colon {\rm continuous \hspace{3pt} functions} \rightarrow {\bf R}$ is linear

(by the Sum Rule and the Constant Multiple Rule), and even

  • the integral as a function $\displaystyle\int \hspace{3pt} \colon {\rm differential \hspace{3pt} forms} \rightarrow {\bf R}$ is linear.