- V is finite dimension means spanned by finite set of vectors - Linearly independent subset - Spanning subset $\equality$ generating subset - Th: number of elements in LI subset is $\lte$ number of elements in spanning set. - Th: - Def: basis satisfies 1,2 - Th; any two bases have same number of elements - Def: dimension is number of elements in basis - Remakr: specify which field: R is subfield of C - notions of linear independence and spanning are different over C and R - {1,i} is basis of C over R - dimension 2 over R; dimension 1 over C - cn over C = n; over R, 2n - Th: every spanning set can be reduced to a basis; in book; - similar to inequality, with n steps; left with step Bn - still spans - and lin indep from lemma from last time: if {u1, ..., um} is linear dependent, then - take largest i labeling the vectors from 1, ...., m; take largest index; may be more; - - insightful lemma; trick: how to argue this thing; - Corollary: 1: consequence of theorem: - spanning subset of m-dimensional subspace of V - to show something is basis, can check both properties: spanning; linearly independent; - now, if know what dimension is, prove something is basis by only checking one property: equals dimension; so only check one thing; don't have to check linear independence - Now, start with Linear independent subset, not redundant, and add to it to be spanning; - Theorem: - Corollary 2: list - Corollary 1 prime; from same theorem 1; every V has basis - Proof: take any finite spanning set, then by Th 1 it contains a basis. - How does basis interact with dimension? interact with subspace - subspace dimension is always lt or equal to dim of V - 48:36 - Theorem 3: if V is fin-dim, then so is U - must prove, since not from axiom: read 2.25 in book - now, skip - now can measure dimensions - Theorem 4; dim U $\lessthan$ or equal to dimension of V - Proof: choose a basis of U - look at it as LI set of U. it is also LI subset of V - LH: linear independent subset of V; RH, spanning subset of V - if two VS - Theorem 5: U is subset or subspace of V, and dim U and Dim V are same then U = V - by Th 2, basis of U is LI. It's in V; by Th 2, LI subset can be extended to basis of V, by adjoining more elements - Div V = dim U - every vector of V is also vector of U; adjoining implies that dim V is bigger than dim - Quantized - Every subspace has a complement such that direct sum of two is V - two 1-dim subspace, if contain each other, are same - Count by integers instead of R; addition ;no mult inverse - multiplication is by ring, not by field Every subspace has a complement. combine the two with direct sum to V - Theorem 6: U inside V. $\exists$ elements, w1, ...., wm; if adjoin to u1, ...um dimU strictly less than dim V; $\oplus$ Finish 2C to finish dimension, span, linear independence - Morphisms; relations between classes or objects - Set theory is simplest: collections of things; already, notion of map - given two sets, have notion of map or function, from one set to another; from s to f(s) - - Category Theory : 109:00 - Set Theory: map, or function is a morphism in the category of sets - Vector spaces have two structures: addition and scalar multiplication - Linear map or linear transformation - - end - Let V, W be two vector spaces over the same field - Def: a linear map from V to W is a map of sets $T: V \rightarrow W$ such that - 1. T (u + v) = T(u) + T(v) $\forall u, v \in V$ - 2 T ($\lambda u$ ) = $\lambda T(u)$ $\forall u \in V, \forall \lambda \in F$ - implications: - Examples - $V = W = \mathbb{R}$ - f $\mathbb{R}\rightarrow \mathbb{R}$ - look at polynomials: $a_nx^n + a_{n-1}x^{n-1} + ... + a_1 x + a_0$ - f (x + y ) = f(x) + f(y) - f($\lambda x$) = $\lambda$ f(x) - f ($\lambda x$) = $\lambda^n x^n$ --- T($0_v$) = $O_w$ seems to be mistake in Theorem proof. To next week