# Differentiation is a Linear Transformation

## Problem 433

Let $P_3$ be the vector space of polynomials of degree $3$ or less with real coefficients.

**(a)** Prove that the differentiation is a linear transformation. That is, prove that the map $T:P_3 \to P_3$ defined by

\[T\left(\, f(x) \,\right)=\frac{d}{dx} f(x)\]
for any $f(x)\in P_3$ is a linear transformation.

**(b)** Let $B=\{1, x, x^2, x^3\}$ be a basis of $P_3$. With respect to the basis $B$, find the matrix representation of the linear transformation $T$ in part (a).

Contents

## Proof.

### (a) Prove that the differentiation is a linear transformation.

Let $f(x), g(x)\in P_3$. By the basic properties of differentiations, we have

\begin{align*}

T\left(\, f(x)+g(x) \,\right)&=\frac{d}{dx}\left(\, f(x)+g(x) \,\right)\\

&=\frac{d}{dx}\left(\, f(x) \,\right)+\frac{d}{dx}\left(\, g(x) \,\right)\\

&=T\left(\, f(x) \,\right)+T\left(\, g(x) \,\right).

\end{align*}

For $f(x)\in P_3$ and $r\in \R$, we also have

\begin{align*}

T\left(\, rf(x) \,\right)&=\frac{d}{dx}\left(\, rf(x) \,\right)\\

&=r\frac{d}{dx}\left(\, f(x) \,\right)\\

&=rT\left(\, f(x) \,\right).

\end{align*}

Therefore, the map $T$ is a linear transformation.

### (b) Find the matrix representation of the linear transformation $T$.

We compute

\begin{align*}

T(1)&=\frac{d}{dx}(1)=0, && T(x)=\frac{d}{dx}(x)=1\\[6pt]
T(x^2)&=\frac{d}{dx}(x^2)=2x, && T(x^3)=\frac{d}{dx}(x^3)=3x^2.

\end{align*}

It follows that the coordinate vectors are

\begin{align*}

[T(1)]_B&=\begin{bmatrix}

0 \\

0 \\

0 \\

0

\end{bmatrix}, && [T(x)]_B=\begin{bmatrix}

1 \\

0 \\

0 \\

0

\end{bmatrix}\\[6pt]
[T(x^2)]_B&=\begin{bmatrix}

0 \\

2 \\

0 \\

0

\end{bmatrix}, && [T(x^3)]_B=\begin{bmatrix}

0 \\

0 \\

3 \\

0

\end{bmatrix}.

\end{align*}

Thus the matrix representation of the linear transformation $T$ is given by

\[\left[\, [T(1)]_B, [T(x)]_B, [T(x^2)]_B, [T(x^3)]_B \,\right]=\begin{bmatrix}

0 & 1 & 0 & 0 \\

0 &0 & 2 & 0 \\

0 & 0 & 0 & 3 \\

0 & 0 & 0 & 0

\end{bmatrix}.\]

Remark that we also write this as

\begin{align*}

\begin{bmatrix}

T(1) &T(x)& T(x^2)& T(x^3)

\end{bmatrix}

\\[6pt]
=\begin{bmatrix}

1 & x & x^2 & x^3

\end{bmatrix}\begin{bmatrix}

0 & 1 & 0 & 0 \\

0 &0 & 2 & 0 \\

0 & 0 & 0 & 3 \\

0 & 0 & 0 & 0

\end{bmatrix}.

\end{align*}

## Related Question.

Let $A$ be the matrix representation obtained in part (b).

It is easy to see that $A^4$ is the zero matrix.

We say that $A$ is a **nilpotent matrix**.

There is a theoretical explanation behind this.

Check out the post “Differentiating Linear Transformation is Nilpotent” for an explanation.

Add to solve later

Sponsored Links

## 1 Response

[…] the post “Differentiation is a linear transformation” for a […]