# 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. Add to solve later

## 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
[T(x)]_B=\begin{bmatrix}
1 \\
0 \\
0
[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. Add to solve later

### 2 Responses

1. 06/13/2017

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