- - - Last time: theorem; Vector space is finite dimensional - This means there exists a finite set of vectors in V such that every other vector can be written as a linear combination of those - w1, ...., wn is the spanning set - u1, ...., um is a set of linearly independent vectors; so, the first one is nonzero - v <= n - diagram - establish and define a basis; spanning set and - linearly independent set of vectors or lists of vectors - - lemma: $\exists$ unique - allows any vector space over a field to be identified to $F^S$ : $S \in F$ ? - - finite-dimensioned spanning set {w1...wn} - linearly independent {u1 ... um} - m $\le$ n number of linearly independent vectors in V =m; spanning set: n; move element from LI to SP; weaker form of inductive proof, since only for finite number of elements, not for infinite sequence of numbers like $N$ - diagram: cells in horizontal array,U: 1, ... m - w: 1....n - lego pieces: move them: - geometric vectors have no coordinate system; choose basis, then have correspondence between vectors and $R^n$ - linear dependence means $a_1x + a_2y$ =0, where the a's are not zero - change basis - ties to homework exercises from last week. This is week of 2B exercises, so last week was 1C and 2A - mistake in coordinate axes with new v1 and v2. change to (1,1) and (-1,1), not (1, -1) --- - trick to show that a basis exists. $\exists$ : start with spanning set, remove any vector that is a linear combination of other vectors; still spanning, but n-1 elements - Jewel: dimension Theorem - any two bases in V have same number of elements: - proof of earlier theorem ;