MIT09.md

9. Independence, Basis, and Dimension

$A_{m\times n}$, m > n ⇒ $\exists$ non-trivial solutions of $AX = 0$

Independence

vectors $\vec v_1, … , \vec v_n$ independent if $\sum\limits_{i=1}^n \lambda_i \vec v_i = 0 \iff \sum\limits_{i=1}^n |\lambda_i| = 0$

Span

span of vectors $\vec v_1, … , \vec v_n \iff$ {$\vec w = \sum\limits_{i=1}^n \lambda_i \vec v_i | \lambda_i \in \mathbb R$}

Basis

vectors $\vec v_1, … , \vec v_n$ basis of space V $\iff$ they linearly independent and span V

Dimension

dimension of space V is number of vectors in basis of V

rank

rg(A) = # pivot columns = dimension of C(A) = dimension of R(A)

dimension of N(A) = # columns - rg(A)