The Range and Kernel of Linear Operators

# The Range and Kernel of Linear Operators

Definition: Let $X$ and $Y$ be linear spaces and let $T : X \to Y$ be a linear operator. The Range of $T$ denoted $\mathrm{range}(T)$ is the image of $X$ under $T$, that is, $\mathrm{range} (T) = T(X)$. The Kernel of $T$ denoted $\ker (T)$ is the set of all points $x \in X$ for which $T(x) = 0$, that is, $\ker (T) = \{ x \in X : T(x) = 0 \}$. |

We will now prove some results regarding the range/kernel of linear operators.

Proposition 1: Let $X$ and $Y$ be linear spaces and let $T : X \to Y$ be a linear operator. Then $\ker (T)$ is a subspace of $X$ and $\mathrm{range} (T)$ is a subspace of $Y$. |

Proposition 2: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces and let $T : X \to Y$ be a bounded linear operator. If $T$ is bounded then $\ker (T)$ is closed in $X$. |

**Proof:**Let $(x_n)$ be a sequence in $\ker (T)$ such that $(x_n)$ converges to some $x \in X$. Then $T(x_n) = 0$ for all $n \in \mathbb{N}$. Since $T$ is bounded (continuous) and since $(x_n)$ converges to $x$ we have that $(T(x_n))$ converges to $T(x)$. But $(T(x_n)) = (0)$. So $T(x) = 0$. Hence $x \in \ker (T)$.

- Alternatively, note that if $T$ is bounded then it is continuous. Since $\{ 0 \} \subset Y$ is closed (as it is a singleton set) we have that $T^{-1}(0) = \ker (T)$ is closed in $X$. $\blacksquare$

For an analogous result to proposition 2, we will see later that The Open Mapping Theorem tells us that if $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ are Banach spaces and $T : X \to Y$ is a bounded linear operator then the range $T(X)$ is closed if and only if $T$ is an open mapping.

Proposition 3: Let $(X, \| \cdot \|_X)$ be a normed linear space and let $f : X \to \mathbb{R}$ be a linear functional on $X$. Then $f$ is a bounded linear functional if and only if $\ker (f)$ is closed. |

**Proof:**$\Rightarrow$ Set $Y = \mathbb{R}$. Then by proposition 2 we have that $\ker (f)$ is closed in $X$.

- $\Leftarrow$ Suppose that $\ker (f)$ is closed and suppose instead that $f$ is not bounded. Then for each $n \in \mathbb{N}$ there exists an $x_n \in X$.

\begin{align} \quad |f(x_n)| \geq n \| x_n \|_X \end{align}

- From our assumption that $\| f \| = \infty$, that is $\| f \| = \sup_{x \in X, \| x \|_X=1} |T(x)|$ we may assume that $\| x_n \| = 1$ for each $n \in \mathbb{N}$, and so:

\begin{align} \quad |f(x_n)| \geq n \quad (*) \end{align}

- Observe that if $f = 0$ then $f$ is bounded. So we may assume that $f \neq 0$. Let $x \in X$ be such that $f(x) \neq 0$,and for each $n \in \mathbb{N}$ let:

\begin{align} \quad s_n = x - \frac{x_n}{f(x_n)} f(x) \end{align}

- Note that each $s_n$ is well-defined since $f(x_n) \neq 0$ by $(*)$. Also observe that $f(s_n) = 0$ for all $n \in \mathbb{N}$. So $(s_n)$ is a sequence in $\ker (f)$. Also note that $s_n$ converges to $x$ since:

\begin{align} \quad \lim_{n \to \infty} \| s_n - x \| = \lim_{n \to \infty} \left \| -\frac{x_n}{f(x_n)} f(x) \right \|_X = \lim_{n \to \infty} \left ( f(x) \frac{\| x_n \|_X}{f(x_n)} \right ) \leq \lim_{n \to \infty} \frac{f(x)}{n} = 0 \end{align}

- So $(s_n)$ is a sequence in $\ker (f)$ that converges to $x \in X$. Since $\ker (f)$ is closed, $x \in \ker (f)$. But $f(x) \neq 0$, which is a contradiction. So the assumption that $f$ is not bounded is false. So $f$ is a bounded linear functional on $X$. $\blacksquare$