Operators as Matrices

One key observation is that numerous functions possess associated Taylor series. We learn from complex analysis that any holomorphic (complex differentiable) function is also analytic, meaning it has a power series expansion.

This holds true for many real-valued functions, which is significant because it allows us to establish a mapping L: ? ? ?^? from the space of functions to a vector space by associating a power series with a vector formed from its coefficients.

Thus, we can link each function to a corresponding coefficient vector.

It's important to note that the addition of two functions directly corresponds to the addition of their coefficient vectors. Here, we are working within an infinite-dimensional vector space, and we will not delve into the rigorous details required for this.

Accepting this mapping allows us to explore what the ordinary differential operator looks like within this vector context.

When we differentiate the function on the left side of the equation and apply the mapping L again, we arrive at:

This leads us to identify a unique matrix that transforms a function's coefficient vector into that of its derivative.

The ellipsis indicates the continuation of this pattern.

Remarkably, as L serves as an isomorphism between vector spaces, we can equate the space of (holomorphic) functions with the vector space of coefficient vectors, thus allowing us to interpret the differential operator as this infinite matrix.

We can even illustrate this with a commutative diagram, but let's remain focused on our current discussion.

Let’s denote this infinite matrix as D.

To solidify this concept, let's consider a small example.

Take the polynomial f(x) = 3x³ + 2x² + x + 5. We can derive its coefficient vector using L and then differentiate it by applying the matrix D from the left. To revert, we can use the inverse of L.

Observe that D², viewed as a matrix product, corresponds to applying differentiation twice to a coefficient vector. This matrix is expressed as:

Generally, we can express this as:

Here, the non-zero entry in the first row is situated in the n+1'th column. Notably, this formula leads us to D? = I, the identity matrix, and incrementing or decrementing the exponent shifts the non-zero diagonal left or right.

Examining D raised to the power of -1 reveals that the n!/0! term vanishes, yielding:

This represents our integral operator with a constant of 0—it is not the inverse of D.

Let’s label this matrix J. We find that DJ = I, but JD ? I. This aligns with the principle that integrating first and differentiating afterward returns us to our original function, as stated by the fundamental theorem of calculus:

d/dx ? f(x) dx = f(x).

However, differentiating first and then integrating results in functions of the form f(x) + c, which is a shifted variant of the original function. It's also clear that the matrix D lacks a two-sided inverse since its determinant is 0.

We are well aware of how to differentiate and integrate polynomials and power series. The truly fascinating developments arise when we create new operators from these. Specifically, let's explore whether we can express the translation operator as a matrix.

The Translation Operator as a Matrix

Surprisingly, we can rationalize the concept of raising a number to an operator. However, we must first clarify what we mean by arithmetic involving operators.

This is straightforward: given operators T and R, which take functions as inputs and produce other functions, we can define their sum and product based on their resultant effect on a function.

(T + R) f(x) = T f(x) + R f(x).

Similarly, we define their product by applying one operator followed by the other:

(TR) f(x) = T (R f(x)).

Now, we can illustrate what it means to raise e to the differential operator d/dx using the power series representation for e^x. Let's attempt this for the ordinary differential operator d/dx, recalling some results from Taylor series.

First, remember that we can express an analytic function f as a power series centered at x as follows:

Next, recall the Taylor expansion of e^x centered at 0:

To apply the aforementioned exotic operator, we must examine its effect on a function:

We utilized the definitions for adding and multiplying operators and incorporated the Taylor series expansion with center x and value z = x + 1.

Thus, we have:

where S f(x) = f(x+1). More broadly, it can be demonstrated using the binomial theorem that:

Equipped with this intriguing outcome, let’s determine the corresponding matrix for this shift operator.

To achieve this, note that D^n/n! can be elegantly expressed using binomial coefficients, and summing these matrices is straightforward as we simply add them entry-wise.

We find:

With our earlier matrix calculations and patterns, we arrive at:

The diagonal consists of 1s, but the underlying pattern is clearer here. Similarly, we can demonstrate that:

How can we utilize this?

This presents a compelling combinatorial structure, as it provides the expansion of a function f(x+a) in terms of its power series centered at 0.

For example, let's apply this to a polynomial:

The matrix e^D can be interpreted as a 3 × 3 matrix in this case, since all other coefficients are 0. We find:

which corresponds to:

This aligns with the binomial theorem, but more generally pertains to power series and not just expansions of functions of the form (x + y)^n = S_y x^n.

We can also express the operator U: f(x) ? f(-x) as the matrix:

Now, we can construct a specific symmetry operator as a combination of these matrices:

I previously discussed this operator in another article:

Reflexive Functions