Matrix Representations for Linear Transformations of the Vector Space of Polynomials

Linear Transformation problems and solutions

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

 

LoadingAdd to solve later

Sponsored Links


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.


LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Linear Algebra
Stanford University Linear Algebra Exam Problems and Solutions
Is an Eigenvector of a Matrix an Eigenvector of its Inverse?

Suppose that $A$ is an $n \times n$ matrix with eigenvalue $\lambda$ and corresponding eigenvector $\mathbf{v}$. (a) If $A$ is...

Close