Matrix Representations for Linear Transformations of the Vector Space of Polynomials

Problem 71

Let $P_2(\R)$ be the vector space over $\R$ consisting of all polynomials with real coefficients of degree $2$ or less.
Let $B=\{1,x,x^2\}$ be a basis of the vector space $P_2(\R)$.
For each linear transformation $T:P_2(\R) \to P_2(\R)$ defined below, find the matrix representation of $T$ with respect to the basis $B$. For $f(x)\in P_2(\R)$, define $T$ as follows.

(a) \[T(f(x))=\frac{\mathrm{d}^2}{\mathrm{d}x^2} f(x)-3\frac{\mathrm{d}}{\mathrm{d}x}f(x)\]

(b) \[T(f(x))=\int_{-1}^1\! (t-x)^2f(t) \,\mathrm{d}t\]

(c) \[T(f(x))=e^x \frac{\mathrm{d}}{\mathrm{d}x}(e^{-x}f(x))\]

 

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Hint.

To find the matrix, we just need to compute $T(1), T(x), T(x^2)$ and read the coefficients of $1, x, x^2$.

Solution.

(a) $T(f(x))=\frac{\mathrm{d}^2}{\mathrm{d}x^2} f(x)-3\frac{\mathrm{d}}{\mathrm{d}x}f(x)$

Since the derivative the constant function $1$ is zero, we have
\[T(1)=0.\] Since the second derivative of the function $x$ is zero, we have
\[T(x)=-3.\] Finally, we calculate
\[T(x^2)=2-6x.\] There with respect to the basis $\{1, x, x^2\}$, the matrix for $T$ is
\[\begin{bmatrix}
0 & -3 & 2 \\
0 &0 &-6 \\
0 & 0 & 0
\end{bmatrix}.\]

(b) $T(f(x))=\int_{-1}^1\! (t-x)^2f(t) \,\mathrm{d}t$

We first compute
\begin{align*}
T(f(x))&=\int_{-1}^{1} \! (t^2-2tx-+x^2)f(t) \, \mathrm{d}t \\
&=\left( \int_{-1}^{1} \! t^2f(t) \, \mathrm{d}t\right)-2\left( \int_{-1}^{1} \! tf(t) \, \mathrm{d}t \right) x+\left(\int_{-1}^{1} f(t) \, \mathrm{d}t \right) x^2.
\end{align*}
Then we have
\begin{align*}
T(1)&=\frac{2}{3} +2x^2 \\
T(x)&=-\frac{4}{x} x\\
T(x^2)&=\frac{2}{5}+\frac{2}{3}x^2
\end{align*}

Here we calculated the integrals
\[\int_{-1}^{1} \! 1 \, \mathrm{d}t= \left[t \right]_{-1}^1=2, \,\,\,\, \int_{-1}^{1} \!t \, \mathrm{d}t=\left[\frac{1}{2}t^2\right]_{-1}^1=0,\] \[\int_{-1}^{1} \! t^2 \, \mathrm{d}t=\left[\frac{1}{3}t^3 \right]_{-1}^1=\frac{2}{3}, \,\,\,\, \int_{-1}^{1} \! t^3 \, \mathrm{d}t=\left[\frac{1}{4}t^4 \right]_{-1}^1=0, \,\,\,\, \int_{-1}^{1} \! t^4 \, \mathrm{d}t=\left[\frac{1}{5}t^5 \right]_{-1}^1=\frac{2}{5}.
\] Therefore, the matrix for the linear transformation with respect to the basis $B=\{1,x,x^2\}$ is
\[\begin{bmatrix}
\frac{2}{3} & 0 & \frac{2}{5} \\[6pt] 0 &-\frac{4}{3} &0 \\[6pt] 2 & 0 & \frac{2}{3}
\end{bmatrix}.\]

(c) $T(f(x))=e^x \frac{\mathrm{d}}{\mathrm{d}x}(e^{-x}f(x))$

 Note that by the product rule for derivatives,
\begin{align*}
T(f(x))&=e^x \left(-e^{-x}f(x)+e^{-x}\frac{\mathrm{d}}{\mathrm{d}x} f(x) \right)\\
&=-f(x)+\frac{\mathrm{d}}{\mathrm{d}x} f(x).
\end{align*}
Thus we have
\begin{align*}
T(1)&=-1\\
T(x)&=-x+1\\
T(x^2)&=-x^2+2x.
\end{align*}
Hence the matrix for the linear transformation $T$ with respect to the basis $B$ is
\[\begin{bmatrix}
-1 & 1 & 0 \\
0 &-1 &2 \\
0 & 0 & -1
\end{bmatrix}.\]

Comment.

This problem hints a connection between linear algebra and calculus, or differential equations.
And yes, they are closely related and differential equations can be solved using techniques of linear algebras.

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