# Differentiating Linear Transformation is Nilpotent

## Problem 453

Let $P_n$ be the vector space of all polynomials with real coefficients of degree $n$ or less.

Consider the differentiation linear transformation $T: P_n\to P_n$ defined by

\[T\left(\, f(x) \,\right)=\frac{d}{dx}f(x).\]

**(a)** Consider the case $n=2$. Let $B=\{1, x, x^2\}$ be a basis of $P_2$. Find the matrix representation $A$ of the linear transformation $T$ with respect to the basis $B$.

**(b)** Compute $A^3$, where $A$ is the matrix obtained in part (a).

**(c)** If you computed $A^3$ in part (b) directly, then is there any theoretical explanation of your result?

**(d)** Now we consider the general case. Let $B$ be any basis of the vector space of $P_n$ and let $A$ be the matrix representation of the linear transformation $T$ with respect to the basis $B$.

Prove that without any calculation that the matrix $A$ is nilpotent.

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Contents

## Background.

### Differentiating is a linear transformation.

Recall that differentiating a polynomial is a linear transformation.

See the post “Differentiation is a linear transformation” for a proof.

The post also contains a similar problem for $n=3$.

### Definition (Nilpotent matrix).

A square matrix $A$ is called **nilpotent** if there exists a positive integer $k$ such that $A^k=O$, the zero matrix.

## Solution.

### (a) Find the matrix representation $A$ of the linear transformation $T$ with respect to the basis $B$.

To find the matrix representation $A$ of $T$, we first find the coordinate vectors as follows.

Since we have

\begin{align*}

T(1)&=0=0\cdot 1+0\cdot x+0\cdot x^2\\

T(x)&=1=1\cdot 1+0\cdot x+0\cdot x^2\\

T(x^2)&=2x=0\cdot 1+2\cdot x+0\cdot x^2

\end{align*}

we have the coordinate vectors of these with respect to the basis $B$ is

\begin{align*}

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

0 \\

0 \\

0

\end{bmatrix}, \quad

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

1 \\

0 \\

0

\end{bmatrix}, \quad

[T(x^2)]_B=\begin{bmatrix}

0 \\

2 \\

0

\end{bmatrix}.

\end{align*}

Hence the matrix representation $A$ of $T$ is

\begin{align*}

A=\begin{bmatrix}

\,[T(1)]_B & [T(x)]_B & [T(x^2)]_B \,\\

\end{bmatrix}

=\begin{bmatrix}

0 & 1 & 0 \\

0 &0 &2 \\

0 & 0 & 0

\end{bmatrix}.

\end{align*}

### (b) Compute $A^3$, where $A$ is the matrix obtained in part (a).

We calculate

\begin{align*}

A^2&=\begin{bmatrix}

0 & 1 & 0 \\

0 &0 &2 \\

0 & 0 & 0

\end{bmatrix}

\begin{bmatrix}

0 & 1 & 0 \\

0 &0 &2 \\

0 & 0 & 0

\end{bmatrix}

=\begin{bmatrix}

0 & 0 & 2 \\

0 &0 &0 \\

0 & 0 & 0

\end{bmatrix} \text{ and },\\[6pt]
A^3&=A^2A=

\begin{bmatrix}

0 & 0 & 2 \\

0 &0 &0 \\

0 & 0 & 0

\end{bmatrix}

\begin{bmatrix}

0 & 1 & 0 \\

0 &0 &2 \\

0 & 0 & 0

\end{bmatrix}

=\begin{bmatrix}

0 & 0 & 0 \\

0 &0 &0 \\

0 & 0 & 0

\end{bmatrix}.

\end{align*}

Thus, $A^3$ is the zero matrix.

### (c) Is there any theoretical explanation of your result?

Recall from calculus that differentiating a polynomial once lowers the degree by one. Thus if we differentiate a degree $2$ polynomial once, then you get a degree $1$ polynomial. If we apply the differentiation again, we obtain a degree $0$ polynomial, which is a constant. If we differentiate once more, it becomes the zero polynomial.

Hence differentiating a polynomial of degree $2$ or less three times, the polynomial becomes the zero polynomial.

Differentiating once is the linear transformation $T$, hence multiplying by $A$ corresponds to differentiating once.

Therefore, $A^3$ corresponds to differentiating three times.

This is why $A^3$ is the zero matirx.

### (d) Prove that without any calculation that the matrix $A$ is nilpotent.

We generalize the argument in part (c) to any positive integer $n$. Recall again differentiating a polynomial once lowers the degree by one.

Thus, if we differentiate a degree $n$ polynomial $n+1$ times, then the polynomial becomes the zero polynomial.

In other words, we have

\[T^{n+1}\left(\, f(x) \,\right)=0\]
for any $f(x)\in P_n$.

Thus we have $T^{n+1}$ is the zero linear transformation.

In terms of the matrix representation, this means that we have $A^{n+1}=O$, the zero matrix.

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