- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.2 Null space and range
$\quad$ In this section, $T$ denotes a linear transformation of a linear space $V$ into a linear space $W.$
$\quad$ Theorem 2.1 $\quad$ The set $T(V)$ (the range of $T$) is a subspace of $W.$ Moreover, $T$ maps the zero element of $V$ onto the zero element of $W.$
$\quad$ Proof. $\quad$ Recall from Theorem 1.4 that if $S$ is a nonempty subset of a linear space $V,$ then $S$ is a subspace if and only if $S$ satisfies the closure axioms.
$\quad$ By definition, if $x$ and $y$ are elements of $V,$ then $T(x)$ and $T(y)$ are elements of $T(V).$ But since $V$ is a linear space, the sum $x + y$ and the scalar product $cx$ (for any scalar $c$) are also elements of $V,$ which means $T(x + y)$ and $T(cx)$ are elements of $T(V).$ But because $T$ is a linear transformation, we have $T(x + y) = T(x) + T(y)$ and $T(cx) = cT(x),$ which shows that for elements $T(x)$, $T(y),$ and any scalar $c,$ the elements $T(x) + T(y)$ and $cT(x)$ are elements of $T(V),$ thus satisfying the closure axioms.
Then, if we let $y = x + (-1)x = O$ be an element of $V$ we have $$T(y) = T(O) = T(x) - T(x) = O,$$ thus mapping the zero element of $V$ onto the zero element of $W. \quad \blacksquare$
$\quad$ Definition. $\quad$ The set of all elements in $V$ that $T$ maps onto $O$ is called the null space of $T$ and is denoted by $N(T).$ Thus, we have: \begin{align*} N(T) &= \{x\, |\, x \in V \ \text{and} \ T(x) = O\}. \end{align*} The null space is sometimes called the kernel of $T.$
$\quad$ Theorem 2.2. $\quad$ The null space of $T$ is a subspace of $V.$
$\quad$ Proof. $\quad$ As a resulf of Theorem 1.4, it will suffice to show that the null space of $T$ satisfies the closure axioms.
$\quad$ Suppose $x$ and $y$ are elements of $N(T)$ (that is, let $x$ and $y$ be elements of $V$ such that $T(x) = O$ and $T(y) = O$) and let $c$ be any scalar. Becasue $N(T)$ is a subset of the linear space $V,$ we know that $x + y$ and $cx$ are also elements of $V.$ To complete the proof, we wish to show that $x + y$ and $cx$ satisfy membership of $N(T).$ In other words, we will show that $T(x + y)$ and $T(cx)$ map to the zero element of $W.$ But because $T$ is a linear transformation, we have: \begin{align*} T(x + y) &= T(x) + T(y) \\ &= O + O \\ &= O \\ \\ T(cx) &= cT(x) \\ &= cO \\ &= O \end{align*} Thus, $x + y$ and $cx$ are elements of $N(T),$ satisfying closure under addition and under multiplication by scalars. This completes the proof.