Question

(calculus required) Define \(T:C\left( {0,1} \right) \to C\left( {0,1} \right)\) as follows: For f in \(C\left( {0,1} \right)\), let \(T\left( t \right)\) be the antiderivative \({\mathop{\rm F}\nolimits} \) of \({\mathop{\rm f}\nolimits} \) such that \({\mathop{\rm F}\nolimits} \left( 0 \right) = 0\). Show that \(T\) is a linear transformation, and describe the kernel of \(T\). (See the notation in Exercise 20 of Section 4.1.)

Short answer

It is proved that \(T\) is a linear transformation. The kernel of \(T\) is \(\left\{ 0 \right\}\).

Step-by-step solution

State the condition for linear transformation

The conditions for linear transformationTare as follows:

1.\(T\left( {{\mathop{\rm u}\nolimits} + {\mathop{\rm v}\nolimits} } \right) = T\left( {\mathop{\rm u}\nolimits} \right) + T\left( {\mathop{\rm v}\nolimits} \right)\) for all \({\mathop{\rm u}\nolimits} ,{\mathop{\rm v}\nolimits} \,\,{\mathop{\rm in}\nolimits} \,\,V\), and

2 \(T\left( {c{\mathop{\rm u}\nolimits} } \right) = cT\left( {\mathop{\rm u}\nolimits} \right)\) for all \({\mathop{\rm u}\nolimits} \,\,\,{\mathop{\rm in}\nolimits} \,\,V\) and all scalar \(c\).

Show that \(T\) is a linear transformation

represents the antiderivative \({\mathop{\rm F}\nolimits} \) of \({\mathop{\rm f}\nolimits} \) with \({\mathop{\rm F}\nolimits} \left( 0 \right) = 0\), and \(T\left( {\mathop{\rm g}\nolimits} \right)\) represents the antiderivative \({\mathop{\rm G}\nolimits} \) of \({\mathop{\rm g}\nolimits} \) with \({\mathop{\rm G}\nolimits} \left( 0 \right) = 0\).

According to the rules of antidifferentiation, \({\mathop{\rm F}\nolimits} + G\) is the antiderivative of \({\mathop{\rm f}\nolimits} + {\mathop{\rm g}\nolimits} \). Then,

\(\begin{array}{c}\left( {{\mathop{\rm F}\nolimits} + G} \right)\left( 0 \right) = \left( {\mathop{\rm F}\nolimits} \right)\left( 0 \right) + \left( G \right)\left( 0 \right)\\ = 0 + 0\\ = 0\end{array}\)

Therefore, \(T\left( {{\mathop{\rm f}\nolimits} + g} \right) = T\left( {\mathop{\rm f}\nolimits} \right) + T\left( g \right)\).

Similarly, \(c{\mathop{\rm F}\nolimits} \) is an antiderivative of \({\mathop{\rm cf}\nolimits} \).

\(\begin{array}{c}\left( {c{\mathop{\rm F}\nolimits} } \right)\left( 0 \right) = c{\mathop{\rm F}\nolimits} \left( 0 \right)\\ = c0\\ = 0\end{array}\)

Therefore, \(T\left( {c{\mathop{\rm f}\nolimits} } \right) = cT\left( {\mathop{\rm f}\nolimits} \right)\), and \(T\) is a linear transformation.

Thus, it is proved that \(T\) is a linear transformation.

Describe the kernel of \(T\)

To determine the kernel of \(T\), find all functions \({\mathop{\rm f}\nolimits} \) in \(C\left( {0,1} \right)\) with the antiderivative equal to the zero function. The zero function (0) is the only function that has this property.

Therefore, the kernel of \(T\) is \(\left\{ 0 \right\}\).