Find the Nullspace and Range of the Linear Transformation $T(f)(x) = f(x)-f(0)$

Linear Transformation problems and solutions

Problem 680

Let $C([-1, 1])$ denote the vector space of real-valued functions on the interval $[-1, 1]$. Define the vector subspace
\[W = \{ f \in C([-1, 1]) \mid f(0) = 0 \}.\]

Define the map $T : C([-1, 1]) \rightarrow W$ by $T(f)(x) = f(x) – f(0)$. Determine if $T$ is a linear map. If it is, determine its nullspace and range.

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$T$ is a lineawr transformation

First we must check that $T$ takes values in $W$, as the problem implies but does not prove. Consider $f \in C([-1 , 1])$. The function $T(f)$ lies in $W$ if and only if $T(f)(0) = 0$. We quickly check that it satisfies this condition:in $C([-1, 1])$ lies
\[ T(f)(0) = f(0) – f(0) = 0 . \]

Now that we know $T$ takes values in $W$, we show that it is a linear transformation. We will prove this by showing that $T$ satisfies both of the axioms for linear transformations. First, suppose that $f, g \in C([-1, 1])$. Then
T(f+g)(x) &= (f+g)(x) – (f+g)(0) \\
&= f(x) + g(x) – f(0) – g(0) \\
&= f(x) – f(0) + g(x) – g(0) \\
&= T(f)(x) + T(g)(x) . \end{align*}

Now for a scalar $c \in \mathbb{R}$ we have
T( cf )(x) &= (cf)(x) – (cf)(0) \\
&= c f(x) – c f(0) \\
&= c ( f(x) – f(0) ) \\
&= c T(f)(x) . \end{align*}

Thus we have proven that $T$ is a linear transformation.

The nullspace of $T$

Next, we will prove that the nullspace of $T$ is
\[\mathcal{N}(T) = \{ f \in C([-1 , 1]) \mid f(x) \mbox{ is a constant function } \}.\] Suppose that $f \in \calN(T)$, that is, $f$ satisfies
\[0 = T(f)(x) = f(x) – f(0).\] Then $f(x) = f(0)$ for all $x \in [-1, 1]$. This means that $f$ is a constant function. On the other hand, if $f(x)$ is a constant function, then $T(f)(x) = f(x) – f(0) = 0$. We see that $f$ lies in the nullspace of $T$ if and only if it is a constant function.

The range of $T$

Next, we want to find the range of $T$. We claim that
\[\mathcal{R}(T) = W.\]

Suppose that $f \in W$, that is, it is a function such that $f(0) = 0$. Then
\[T(f)(x) = f(x) – f(0) = f(x),\] and so we see that $f \in \mathcal{R}(T)$.
Conversely, suppose that $f \in \mathcal{R}(T)$, so that $f = T(g)$ for some $g \in C([-1, 1])$. Then
\[f(0) = T(g)(0) = g(0) – g(0) = 0,\] and so $f \in W$. Thus every $f \in C([-1, 1])$ lies in the range of $T$ if and only if $f(0) = 0$.


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