# 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).

## 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.

### 1 Response

1. 06/13/2017

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

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