# Showing differentiation is a linear map

• HmBe
In summary: So in summary, the conversation is about a problem involving differentiation and linear transformations. The conversation covers the correct matrix representation for part ii) of the problem and discusses the requirements for a linear transformation to be fulfilled in order to show the matrix representation for part i). The conversation also includes a clarification on the matrix representation for part ii) and provides an example to demonstrate its correctness.
HmBe

## The Attempt at a Solution

For part ii) I wrote it out as a matrix, getting

\begin{array}{ccccccc}
0 & 0 & 0 & 0 & ... & 0 \\
0 & 0 & 2 & 0 & ... & 0 \\
0 & 0 & 0 & 6 & ... & 0 \\
. & . & . & . & . & . \\
0 & 0 & 0 & 0 & ... & N(N-2) \end{array}

So the rank = N-2 and the nullity = 2

However for part i) I'm not entirely sure how to go about it. Is it enough to show the map can be written as a matrix form, or do I need to show the additivity and scalar multiplication things hold over differentiation? Or is either ok?

Cheers.

My guess is that the teacher is expecting you to show the explicit requirements for a linear transformation are fulfilled, for which you don't need to use matrices at all.

But once you have shown that a matrix can be used (formally proving this is probably pretty hard), the first part is implied.

The teacher probably doesn't want to you prove from first principles that differentiation is linear (unless this is an upperlevel analysis course in disguise). Just show briefly that the known rules about derivatives imply that they are linear. At least that is what my prof. wanted from a similar question when I took LA

Thanks.

I've kinda got another question though. I thought I understood the matrix for part ii, but thinking about it it doesn't really make much sense. Can anyone help me understand it?

Cheers.

First of all, the matrix isn't quite right, but I didn't notice that the first time I looked at it. That may be your only issue, but I'll be explicit about everything in case something else is confusing you.

Does it make sense that you can view polynomials in a vector space? Just like in a vector space, the coefficients of any single term in the polynomial do not effect the other terms. If you have $5x^3$, the 5 only effects the "$x^3$ direction". When you add up all of the terms, the number of $x^3$ added in will be 5 no matter how many times $x^2$ you also add in. Yes, when you evaluate there may be different values of $x$ such that the polynomial equals the same thing, but we are not dealing with final value when we differentiate. The map for differentiation is,
$D: \mathbb{R}[x]\rightarrow\mathbb{R}[x]$ Hopefully that part makes sense. If not, make sure that you are comfortable with the idea that polynomials fit the requirements to be a vector space.

When you differentiate a single power of $x$, the exponent is multiplied by the coefficient and then that value (coeff x exponent) is moved into the position as coefficient for one power down of $x$. In calc you think about this as subtracting one from the exponent, but here you are being asked to imagine that all (up to N) of the possible values of $x^n$ exist, but may be multiplied by zero the same way that a vector point down the $x-$axis doesn't mean that $y\mbox{ and } z$ disappear, they just have zeros for coefficients.

So our N+1 dimensional vector $[c_0,c_1,c_2,\ldots,c_N]^T$ of the coefficients of the polynomial $c_0x^0+c_1x^1+c_2x^2+\ldots+c_Nx^N$ can be written as a column vector with $c_0$ at the top and $c_N$ at the bottom. When you write the matrix for $D$, it puts $c_1$ in the $c_0$ place and gets rid of $c_0$. That is just the normal rule from calculus. Then $c_n*n\rightarrow c_{n-1}$ follows in general so the matrix will have 1,2,3,4... just off the diagonal

$\begin{array}{cccccc} 0&1&0&0&\ldots&0\\ 0&0&2&0&\ldots&0\\ 0&0&0&3&\ldots&0\\ \ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\ 0&0&0&0&\ldots&N\\ 0&0&0&0&\ldots&0\\ \end{array}$

When you square this matrix, you will get$\begin{array}{cccccc} 0&0&2&0&\ldots&0\\ 0&0&0&2*3&\ldots&0\\ \ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\ 0&0&0&0&\ldots&N(N-1)\\ 0&0&0&0&\ldots&0\\ 0&0&0&0&\ldots&0\\ \end{array}$

Which is the same as taking the derivative twice. If you are not convinced, try an easy polynomial like $1+x+x^2+x^3+x^4+x^5$ using your normal calc rules and then do it with a matrix. Then you can put the $c_n$ in as coefficients to finally remove all doubt.

Last edited:
Yeah, that's what I got. Except I got N(N-1) rather then (N-1)(N-2).

Thanks a lot for the help.

Yes, that is a typo. sorry.

## 1. What is "showing differentiation is a linear map"?

"Showing differentiation is a linear map" refers to the process of proving that the derivative operator, which maps a function to its derivative, is a linear transformation. This means that it satisfies the properties of linearity, such as preserving addition and scalar multiplication.

## 2. Why is it important to show that differentiation is a linear map?

Proving that differentiation is a linear map is important because it establishes a fundamental property of the derivative operator. This allows us to use linear algebra techniques to solve differential equations and other problems involving derivatives.

## 3. How is differentiation shown to be a linear map?

Differentiation can be shown to be a linear map by using the definition of linearity, which states that a transformation is linear if it satisfies the properties of preservation of addition and scalar multiplication. In the case of differentiation, these properties can be demonstrated using the properties of derivatives.

## 4. What are the applications of showing differentiation is a linear map?

Showing differentiation is a linear map has many applications in mathematics and other fields. It allows us to use linear algebra to solve problems involving derivatives, such as optimization and curve fitting. It is also the basis for many important theorems in calculus, such as the chain rule and the fundamental theorem of calculus.

## 5. Can differentiation ever not be a linear map?

No, differentiation is always a linear map. This is because it satisfies the properties of linearity, which are inherent in the definition of the derivative. However, there are cases where the derivative may not exist or may be undefined, such as at points where the function is not differentiable.

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