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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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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