- 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
3.9 $\quad$ Determinants and independence of vectors
$\quad$ We can use Theorem 3.5 to deduce a simple criterion for the independence of vectors:
$\quad$ Theorem 3.6. $\quad$ A set of $n$ vectors $A_1, \dots, A_n$ in $n$-space is independent if and only if $d(A_1, \dots, A_n) \neq 0.$
$\quad$ Proof. $\quad$ In Theorem 3.1 (e), we proved that if $A_1, \dots, A_n$ are dependent, then $d(A_1, \dots, A_n)= 0.$ Thus, $d(A_1, \dots, A_n) \neq 0$ implies that $A_1, \dots, A_n$ are independent. To prove the converse, we assume that $A_1, \dots, A_n$ are independent and show that this implies $d(A_1, \dots, A_n) \neq 0.$
$\quad$ If $A_1, \dots, A_n$ are independent vectors in $V_n,$ then they form a basis for $V_n.$ Then, by Theorem 2.12, if $e_1, \dots, e_n$ are the unit coordinate vectors of $V_n,$ there is one and only one linear transformation $T: V_n \to V_n$ such that \begin{align*} T(A_k) &= e_k, \qquad \text{for $k = 1, \dots, n.$} \end{align*} Then, if we set $B$ as the matrix representation of $T,$ setting $A = \begin{bmatrix}A_1 \\ \vdots \\ A_n\end{bmatrix},$ we find that $BA = I.$ But this means that $A$ is nonsingular. But by Theorem 3.5, this means that $\det A = d(A_1, \dots, A_n) \neq 0. \quad \blacksquare$