- 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
2.4 Exercises
30. $\quad$ Let $T:V\to W$ be a linear transformation of a linear space $V$ into a linear space $W.$ If $V$ is infinite-dimensional, prove that at least one of $T(V)$ or $N(T)$ is infinite-dimensional.
[Hint: Assume $\dim N(T)=k,$ $\dim T(V)=r,$ let $e_1,\dots,e_k$ be a basis for $N(T)$ and let $e_{k+1},\dots,e_{k+n}$ be independent elements in $V,$ where $n>r.$ The elements $T(e_{k+1}),\dots,T(e_{k+n})$ are dependent since $n>r.$ Use this fact to obtain a contradiction.]
Proof. $\quad$ If the set $\{T(e_{k+1}), \dots, T(e_{k+n})\}$ is dependent, then by definition there exists some combination of scalars $a_1, \dots, a_n,$ not all zero, such that \begin{align*} \sum_{i = 1}^n a_i T(e_{k + i}) &= O \end{align*} But because $T$ is a linear transformation, this relation is equivalent to \begin{align*} T\left(\sum_{i = 1}^n a_i e_{k + i}\right) &= O \end{align*} This means that the element $v = \sum_{i = 1}^n a_i e_{k + i}$ is in the null space of $T.$ But, $\{e_1, \dots, e_k, e_{k+1}, \dots, e_{k+n}\}$ is independent. Hence, if $v$ were in the span of $\{e_1, \dots, e_k\},$ it would mean that there exists a nontrivial linear combination of $e_1, \dots, e_k, e_{k+1}, \dots, e_{k+n}$ equal to $O,$ making the set dependent and contradicting our given information. Thus, since $v$ is a nonzero element of $N(T)$ not in the span of $\{e_1, \dots, e_k\},$ then \begin{align*} \sum_{i=1}^{k}a_i e_i + bv = O \end{align*} implies that the scalars $a_1, \dots, a_k, b$ are all zero. In other words, $\{e_1, \dots, e_k, v\}$ is an independent set of elements in $N(T).$ From Theorem 1.7, this implies that $\{e_1, \dots, e_k, v\}$ is a subset of the basis of $N(T),$ making $\dim N(T) \geq k+1,$ leading to a contradiction.
$\quad$ If we take another $n$ elements $e'_{1}, \dots, e'_{n}$ in $V$ such that the set \begin{align*} \{e_1, \dots, e_k, v, e'_{1}, \dots, e'_{n}\} \end{align*} is independent (we can always find $n$ such elements since $V$ is infinite-dimensional), then because $n \gt \dim T(V),$ the elements \begin{align*} T(e'_1), \dots, T(e'_n) \end{align*} of $T(V)$ are dependent. By the process used before, this implies that the element $$v_1 = \sum_{i = 1}^{n}b_ie'_i$$ of $V$ (where $b_1, \dots b_n$ are scalars) is also in the basis of $N(T).$ By induction, this shows that if $\dim T(V)$ is finite, then $\dim N(T)$ must be infinite. Otherwise, if $\dim N(T)$ remains finite, then by the Nullity Plus Rank Theorem, $\dim T(V)$ must be infinite. As such, we have shown that if $V$ is infinite-dimensional, then at least one of $N(T)$ or $T(V)$ is infinite-dimensional. This completes the proof.