- 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
2.3 Nullity and rank
$\quad$ In this section, as before, $T$ denotes a linear transformation of a linear space $V$ into a linear space $W.$ If $V$ is finite-dimensional, then the null space is also finite dimensional since $N(T)$ is a subspace of $V.$ The dimension of $N(T)$ is called the nullity of $T.$ In the next theorem, we will prove that the range of $T$ is also finite-dimensional, with this dimension being called the rank of $T.$
$\quad$Theorem 2.3. $\quad$ Nullity plus rank theorem. $\quad$ If $V$ is finite-dimensional, then $T(V)$ is also finite-dimensional and we have \begin{align*} (2.1) &\qquad \dim N(T) + \dim T(V) = \dim V \end{align*} In other words, the nullity plus the rank of a linear transformation is equal to the dimension of its domain.
$\quad$ Proof. $\quad$ Let $n$ be the dimension of $V$ with $e_1, \dots, e_k$ ($k \leq n$) being a basis for $N(T).$ Then, by Theorem 1.7, these elements are part of some basis \begin{align*} (2.2) &\quad e_1, \dots, e_k, e_{k+1}, \dots e_n \end{align*} for $V.$ Now, let $v$ be some element in $V.$ We have: \begin{align*} v &= \sum_{i=1}^n a_ie_i \end{align*} whose linear transformation $T(v)$ is: \begin{align*} T(v) &= \sum_{i=1}^n a_iT(e_i) \end{align*} But because $N(T)$ is spanned by $e_1, \dots, e_k,$ this means that for any $k,$ scalars $a_1, \dots, a_k,$ we have: \begin{align*} \sum_{i=1}^k a_iT(e_i) = O. \end{align*} In other words, $T(V)$ is spanned by the $n - k$ components \begin{align*} (2.3) &\quad T(e_{k+1}), \dots, T(e_n). \end{align*} To show that $T(e_{k+1}), \dots, T(e_n)$ forms a basis for $T(V),$ we must show that the set $\{T(e_{k+1}), \dots, T(e_n)\}$ is independent. Suppose there are $n - k$ scalars such that \begin{align*} \sum_{i= k+1}^n c_iT(e_i) &= O \end{align*} then, because $T$ is a linear transformation, we have: \begin{align*} T\left(\sum_{i= k+1}^n c_ie_i\right) &= O, \end{align*} which means that $x = c_{k+1}e_{k+1} + \cdots + c_n$ is an element of the null space. As such, there are $k$ scalars $c_1, \dots, c_k$ such that $x = c_1e_1 + \cdots + c_ke_k,$ giving us: \begin{align*} x - x &= \sum_{i= 1}^k c_ie_i\ - \sum_{i= k+1}^n c_ie_i\ = O. \end{align*} But $e_1, \dots, e_n$ form a basis for $V,$ thus they are independent elements of $V,$ which means that for the above equation to hold, $c_1, \dots, c_n$ must all be zero. But this implies that the elements of $(2.3)$ are also independent, making $\{T(e_{k+1}), \dots, T(e_n)\}$ a basis for $T(V),$ with $\dim T(V) = n - k$ and $$ \dim N(T) + \dim T(V) = \dim V. \quad \blacksquare$$
Note: If $V$ is infinite-dimensional, this means that at least one of $N(T)$ or $T(V)$ is infinite-dimensional.