Lecture 6: Corrections; Dimension; Linear maps Travis Schedler Tues, Sep 28, 2010 (version: Tues, Sep 28, 4:45 PM)
Goal To briefly correct the proof of the main Theorem from last time. (See website for details) To understand dimensions of subspaces and sums To introduce linear maps. Motivation: Understand relationships between vector spaces!
The fundamental inequality corrected Theorem (Theorem 2.6) The length of every linearly independent list is less than or equal to the length of every spanning list. I made a mistake in the proof last time: Given a lin. ind. list (u 1,..., u m ) and a spanning list (v 1,..., v n ). Idea: want {u 1,..., u m } {v 1,..., v n }. Last time I said: Make all the nonzero v j appear in (u 1,..., u m ). Problem: can t if the (v 1,..., v n ) are lin. dependent (and none are zero). Solution one: Can first replace (v 1,..., v n ) by a basis. Better: Just replace any u k / {v 1,..., v n } by any v j / Span(u 1,..., u k 1, u k+1,..., u m ). Eventually, {u 1,..., u m } {v 1,..., v n }. Since they are distinct, m n.
Corollary (Corollary 2.11 + more) Every finite-dimensional vector space V has a basis, and every basis has length dim V. Proof. Every spanning list of length dim V must be minimal, hence linearly independent, i.e., a basis. As explained last time, the theorem implies that all bases have the same length. Corollary (Theorem 2.12) Every linearly independent list in a finite-dimensional vector space can be extended to a basis. Proof. Linearly independent lists in V of length < dim V don t span, so can be extended. Linearly independent lists of length dim V must have length exactly dim V. Thus, they are maximal and span. If a linearly independent list already has length dim V, it is a basis.
Extending bases vs. direct sums Theorem Given any subspace U V, (i) Any basis (u 1,..., u m ) of U admits an extension (u 1,..., u m, w 1,..., w n ) to a basis of V. (ii) There exists a subspace W V such that V = U W. Proof: For (i), since (u 1,..., u m ) is linearly independent, it extends to a basis. For (ii), let W := Span(w 1,..., w n ).
Dimension and Subspaces Theorem (Propositions 2.7 and 2.15) If V is finite-dimensional and U V is a subspace, then U is finite-dimensional, and dim U dim V. Proof. Any linearly independent list in U is also linearly independent in V, so has length at most dim V. So, there is a maximal such list, which must be a basis. Since it has length at most dim V, we obtain that dim U dim V.
Infinite-dimensionality Conversely: If U V and U is infinite-dimensional, then so is V. Examples: P(R) R Cont. fns. R R All functions R R. So all infinite-dimensional! F n P(F), by polynomials of degree n 1. So dim(p(f)) n for all n, i.e., infinite-dimensional.
Dimension arithmetic Theorem (Theorem 2.18) Let U, W V. Then, dim(u + W ) = dim U + dim W dim(u W ). Proof. Take a basis of U W, (v 1,..., v k ), k = dim(u W ). Extend it to a basis of U: (v 1,..., v k, u 1,..., u m ), k + m = dim U. On the other hand, extend (v 1,..., v k ) to a basis of W : (v 1,..., v k, w 1,..., w n ), k + n = dim W. Thus, (u 1,..., u m, v 1,..., v k, w 1,..., w n ) spans U + W. Claim: it is lin. ind., hence a basis. This implies dim(u + W ) = m + k + n = (m + k) + (n + k) k = dim U + dim W dim(u W ).
More on dimensions Proposition (Proposition 2.19) Suppose V is finite-dimensional and U 1,..., U m V are subspaces such that V = U 1 + + U m. Then, this sum is direct if and only if dim(v ) = dim(u 1 ) + + dim(u m ). (1.2) Proof. In the case m = 2, this was (almost) the warmup to last class! Generally, we show that the sum is direct if and only if, given bases of U 1,..., U m, put together they give a basis of V. Since, put together, they always span, they give a basis if and only if the size, dim U 1 + + dim U m, equals dim V.
Linear maps Motivations: Explain the inclusions of polynomials, continuous functions, all functions, etc. Understand more general linear maps: Projections onto planes and lines Rotations, reflections Fourier transforms More general changes of bases Systems of equations!
Definition of linear maps Definition A linear map is a function T : V W between vector spaces V and W preserving addition and scalar multiplication. Explicitly, T is a function satisfying: T (u + v) = T (u) + T (v), for all u, v V, T (λv) = λt (v). Definition The set of linear maps V W is denoted L(V, W ). We will prove shortly that this is itself a vector space (and more).
Basic examples of linear maps The zero map 0 : V W : 0(v) = 0 for all v V. The identity map I : V V : I (v) = v for all v. Inclusions of subspaces: T : U V where U V. Here T (u) = u. Scalar multiplication: λi : V V : λi (v) = λv for all v. (Can combine with above!)
More sophisticated examples Differentiation of differentiable functions (R to R). E.g., x n nx n 1. Gives P(R) P(R), linear. Integration of continuous functions (R to R): b Definite integration: given a b, f f (y) dy R. Gives a C(R) R. x Indefinite integration: f f (y) dy, gives C(R) C(R). a Multiplication by a function: Fix a function f. This defines M f : functions functions, M f (g) := fg. For example, multiplication by 2x 3 : M 2x 3(1 + x) = 2x 3 + 2x 4. Backward shift: T : F F, T (x 1, x 2, x 3,...) = (x 2, x 3,...). Projections: Given V = U W, we obtain a map V U called projection: (u + w) u. Since all v can uniquely be written as v = u + w, the map v w is well-defined!