Derivative as a matrix: $\mathbf{D}=\dfrac{\mathrm{d}}{\mathrm{d}x}$

Question

I have a strange question, it is possible to consider the derivative as a matrix? (Both are linear transformation technically).
I thought about this example, since $1, x,x^2,...,x^n$ can be thought of as the basis of a vector space, I can consider a polynomial as a vector of its coefficients:
Let $p_n(x)=a_0+a_1 x+a_2 x^2+...+a_n x^n$
Its derivative is $p'_n(x)=\dfrac{\mathrm{d}}{\mathrm{d}x}p_n(x)=a_1+2a_2 x+...+n a_n x^{n-1}$
(and so far nothing new)
But if I consider $p_n(x)$ as a vector of his coefficients: $(a_0,a_1,...,a_n)$ and I want to create a matrix that transforms $p_n(x)$ into $p'_n(x)$ I have: $$\begin{pmatrix}0&1&0&\cdots&0\\ 0&0&2&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&0\\ 0&0&0&\cdots&n\\ 0&0&0&\cdots&0 \end{pmatrix}\begin{pmatrix}a_0\\ a_1\\ a_2\\ \vdots\\ a_n\end{pmatrix}=\begin{pmatrix}a_1\\ 2a_2\\\vdots\\ n a_n\\ 0\\ \end{pmatrix}$$ So technically $\mathbf{D}=\begin{pmatrix}0&1&0&\cdots&0\\ 0&0&2&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&0\\ 0&0&0&\cdots&n\\ 0&0&0&\cdots&0 \end{pmatrix}$ represents the derivative.
After this I tried to do some operations on it, and it came out that:

1. $\mathbf{D}^n$ represents $\dfrac{\mathrm{d}^n}{\mathrm{d}x^n}$
2. $\det(\mathbf{D})=0$, this means that $\mathbf{D}^{-1}$ is not defined (I interpreted this fact as the fact that the inverse operation of the derivative is the integral, which in general is not unique since there are infinitely many that vary for arbitrary constants, EVEN IF it is possible to calculate the pseudoinverse and it gives the integral)
3. $\text{trace}(\mathbf{D})=0$ (I don't know how to interpret that), idem for $\mathbf{D}^{\top}$
4. In general, now I was working on a polynomial, so the dimension of the matrix is $n\times n$ with rank $n-1$, but for a general function I suppose it is $"\infty\times\infty"$ (I don't know if it can be defined the rank for an infinite matrix)
5. $\exp(\mathbf{D})$ gives the upper Pascal Matrix
6. $\mathbf{D}=\text{diag}(1,1,2,...,n!)^{-1}\begin{pmatrix}0&1&0&\cdots&0\\ 0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&0\\ 0&0&0&\cdots&1\\ 0&0&0&\cdots&0 \end{pmatrix}\text{diag}(1,1,2,...,n!)$
   Are these things correct? I'm curious to see if it also has applications for partial derivatives or fractional calculus.
   (I tried to search on the internet "derivative as a matrix" but the main result was the Jacobian, so tell me if this has a name)

Answer 1

Broadly speaking, yes. The derivative is a linear operator, meaning that $\frac{d}{dx}\left(a f(x) + b g(x)\right) = a \frac{d}{dx} f(x) + b \frac{d}{dx} g(x)$, as long as you are working in a vector space where the vectors are differentiable functions.

If the vector space has finite dimension, then it is isomorphic to either $\mathbb{R}^n$ or $\mathbb{C}^n$ depending on which base field we're using, and we can always relate linear transformations in finite dimensional vector fields to matrices in the corresponding $\mathbb{F}^n$, and we can construct that matrix by mapping a basis of the vector space to the standard basis vectors and looking at how the transformation acts on each element of the basis.

So, for example, the vector space of polynomials of degree $n$ or less has a basis $\{1, x, x^2, \ldots, x^n\}$, and you get the results you've already found. You could similarly look at the space of functions spanned by $\{\cos x, \sin x, e^x, e^{-x}\}$ (which happens to be the space of solutions to $y'''' = y$), in which case you'd find that differentiation looks like:

$\frac{d}{dx} \cong \begin{pmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$

and notice that this matrix is invertible, because in this particular space there are no constant functions (other than zero).

You can also look at integration as a linear transformation, but it's a little trickier. You can look at specific definite integrals, e.g. $\int_0^1 f(x) dx$, or definite integrals with variable indices, e.g. $\int_0^x f(t) dt$, and consider how different types of functions get mapped (you'll find that the codomain is no longer the same space, so take that into account).

However, when you have an infinite-dimensional vector space (such as "the space of all polynomials", or "the space of all analytic functions"), you lose the idea of being able to map linear transformations to matrices. So while differentiation remains a linear transformation and you can apply a lot of linear algebra theory to it, you can't rely on any results that assume a finite basis.

Answer 2

This question is not strange at all, and your observations are basically correct.

If a vector space $\mathcal F$ of functions is closed under the derivative (that is, for every $f \in \mathcal F$, $f' \in \mathcal F$), then the derivative is a linear transformation $\mathcal F \to \mathcal F$, precisely because for $f, g \in \mathcal F$ and scalars $c, d$ we have $$(c f + d g)' = c f' + d g'$$ (naturally, this property is called linearity).

Vector spaces $\mathcal F$ of functions closed under the derivative include:

• the vector space $\mathcal P_n$ of polynomials on $\Bbb R$ of degree $\leq n$ for any $n$,
• the vector space $\mathcal P$ of all polynomials on $\Bbb R$,
• the vector space $C^\omega(\Bbb R)$ of all analytic functions on $\Bbb R$, and
• the vector space $C^\infty(\Bbb R)$ of all smooth (i.e., infinitely differentiable) functions on $\Bbb R$.

As in the question, it's convenient to denote the derivative by a symbol like $\bf D$, so that we can write the derivative of a function $f \in \mathcal F$ in the usual notation we use when applying linear transformations, i.e., $${\bf D}f := f' .$$ By definition, applying the derivative $k$ times gives the $k$th derivative, i.e., $${\bf D}^k f = f^{(k)} .$$

For all of the above examples ${\bf D}$ is not invertible, simply because $1 \in \mathcal F$ and ${\bf D}(1) = 0$; in fact $\ker {\bf D} = \operatorname{span}\{1\}$, so $\dim \ker {\bf D} = 1$.

By definition any eigenvector of $\bf D$, say, of eigenvalue $\lambda$, is a function $f$ satisfying $f' = \lambda f$ for some $f$, so it must have the form $f(x) = C e^{\lambda x}$. When we regard the elements of a vector space as functions, sometimes we refer to the eigenvectors as eigenfunctions.

The (finite-dimensional) case $\mathcal F = \mathcal P_n$. For any polynomial $p(x)$ of degree $\leq n$ we have ${\bf D}^{(n + 1)} (p) = p^{(n + 1)} = 0$, so in this case $\bf D$ is nilpotent of order at most $n + 1$, and since $\dim \ker {\bf D} = 1$, $\bf D$ is nilpotent of order exactly $n + 1$. So, the Jordan normal form representation of ${\bf D}$ consists of a single $n \times n$ Jordan block $J$ of eigenvalue $0$. (When $\dim \mathcal F = \infty$ and the domain of the functions in $\mathcal F$ is connected, ${\bf D}$ is not nilpotent---illustrating that some care must be taken when extending our analysis to the case that $\dim \mathcal F = \infty$.)

For any basis of $\mathcal P_n$, we can compute the matrix representation $[{\bf D}]$ of ${\bf D}: \mathcal P_n \to \mathcal P_n$. For the standard basis, $\{1, x, \ldots, x^n\}$, $[{\bf D}]$ is the matrix in the question statement. A natural basis for which $[{\bf D}] = J$ is $$({\bf D}^n (x^n), {\bf D}^{n - 1} (x^n), \ldots, {\bf D}(x^n), x^n) = \left(n!, n! x, \ldots, n x^{n - 1}, x^n\right) .$$ Many other natural bases are produced by starting with the standard basis and applying the Gram-Schmidt algorithm to construct a basis orthonormal with respect to a given inner product on $\mathcal P_n$ (this comment applies just as well to $\mathcal P$).

Your observation (5) that with respect to the standard basis $\mathcal B$ the exponential $[\exp {\bf D}] = \exp [{\bf D}]$ has matrix representation the "upper Pascal matrix", i.e., that $$[\exp {\bf D}]_{ij} = {j \choose i},$$ says (via the Binomial Theorem) that for all $j$, $$(\exp {\bf D})(x^j) = \sum_{i = 1}^j {j \choose i} x^i = (x + 1)^j,$$ and hence by linearity $$(\exp {\bf D})(p)(x) = p(x + 1)$$ ---which in particular furnishes a basis-independent interpretation of $\exp {\bf D}$. More generally, we have $$(\exp t {\bf D})(p)(x) = p(x + t) .$$ (In fact, this statement also applies even when $\dim \mathcal F = \infty$, at least if $\mathcal F \subseteq C^\omega(\Bbb R)$.)

Finally, interpreting the trace of a matrix is not as easy as interpreting its determinant, but $$\left.\frac{d}{dt} \det \exp t {\bf D} \right\vert_{t = 0} = \operatorname{tr} {\bf D} = 0,$$ that is, $\operatorname{tr} {\bf D} = 0$ (observation (3)) implies that the map $p(x) \mapsto p(x + t)$ preserves any volume form on $\mathcal P_n$ to first order in $t$. In fact, since ${\bf D}$ is nilpotent, $\det \exp t{\bf D} = 1$ for all time $t$, that is, $p(x) \mapsto p(x + t)$ preserves any volume form for all time $t$.

Notice, by the way, that even for finite-dimensional $\mathcal F$ the linear transformation ${\bf D}$ need not be singular, let alone nilpotent. For example, if we take $\mathcal F = \Bbb R \cdot \exp = \{x \mapsto c \exp x \mid c \in \Bbb R\}$, ${\bf D}$ is the identity transformation.

Answer 3

After working with vector spaces in linear algebra and getting an idea of how transformation are applied we all apparently think of vectors while thinking of these stuff but cool part is this all can be extented and it is extented to many other spaces called as Abstract Vector spaces.

It take a bit to wrap your head around the fact that all ther associated terms such as spaces subspaces linear transformation are not just "vector", they can be functions series and much more.

Anyway we define a space by few main properties: addition (that adding them together an applying transformation is same as applying transformation then adding each of them) and scalar multiplication (the multiplying by scalar and then applying transformation is same as first applying transformation then scaling).

If any functions series vectors satisfy the very basic property of addition and scalar multiplication they defines a space which may not be vectors.

Hope this helps a bit!