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).
Sponsored Links
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.

Sponsored Links
1 Response
[…] the post “Differentiation is a linear transformation” for a […]