ELEMENTARY LINEAR ALGEBRA

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1 ELEMENTARY LINEAR ALGEBRA DAVID MEREDITH These notes are about the application and computation of linear algebra objects in F n, where F is a field. I think everything here works over any field including finite fields, not just R and C. 1. Introduction: what are the questions to be answered? (1) Important computational questions in linear algebra (a) How can you construct subspaces? (b) How do you decide if a vector is in a subspace? (c) How can you generate vectors in a subspace? (d) How can you decide if one subspace is contained in another? (e) How do you determine the dimension of a subspace? (2) You have at least four ways of creating subspaces (a) the span of a set of vectors (b) the solutions to a set of homogeneous linear equations (c) the sum of subspaces (d) the intersection of subspaces (3) If you define subspaces with spanning sets, what questions are easy to answer? (a) You can generate vectors in the subspace (b) You can find the sum of subspaces: span(s 1 ) + span(s 2 ) span(s 1 S 2 ) (4) If you define subspaces with spanning sets, what questions are hard to answer? (a) Is a vector in the subspace? (b) Finding the intersection of subspaces (5) If you define subspaces with equations, what questions are easy to answer? (a) Is a vector in the subspace? (b) Finding the intersection of subspaces: just take all the equations defining the components of the intersection. (6) If you define subspaces with equations, what questions are hard to answer? (a) How can you generate vectors in a subspace? (b) How can find the sum of subspaces: span(s 1 ) + span(s 2 ) span(s 1 S 2 ) (7) No matter how you define a subspace, it seems hard to find its dimension! (8) We could solve all problems but the dimension problem if we could find a spanning set whenever we had equations and find equations whenever we were given a spanning set. Date: September 5,

2 2 DAVID MEREDITH (a) To determine if U V, find a finite spanning set for U and equations for V, then determine if the spanning vectors satisfy the equations. (9) We could solve the dimension problem if we could find a basis for a subspace. (10) There are constructive algorithms for solving all these problems. We will focus on two algorithms. 2. Vectors and equations F n has the wonderful property that its elements are also functions on the elements. An n-tuple (a 1,..., a n ) where a i F, can be thought of as a point in F n or as a linear function f : F n F defined by f(x 1,..., x n ) a 1 x a n x n. (1) An n-tuple defines a function by the dot product. (2) A system of homogeneous linear equations has the form for some vectors v 1,..., v t F n. v 1 x 0. v t x 0 3. Bases with pivots The main idea in this section is a generalization of the concept of row-reduced matrices (1) A set of vectors in F n is a basis with pivots if each vector has a coordinate equal to 1 in a position where all the other vectors have 0. (2) The distinguished 1 s are called pivots. Each vector has one. The positions where the pivots appear are the pivot positions. (3) The maximum number of vectors in a basis with pivots is n. If there are n vectors, then they are the elementary basis vectors. (4) A basis with pivots is a linearly independent set Proof. Let the vectors be v 1,..., v t with pivot positions c 1,..., c t. That is v i [c j ] δ ij Suppose t a kv k 0. We must show a i 0 for 1 i t. Considering each coordinate, t a kv k [j] 0 for 1 j n. In particular, for each 1 i t, 0 t a kv k [c i ] t a kδ ik a i. (5) Given any finite set of vectors spanning a subspace, you can use row reduction to find a basis with pivots spanning the same subspace. Proof. (a) Row reduction applied to a set of vectors uses three kinds of operations, each reversible and each not changing the span of the vectors involved (i) Change a vector by adding a multiple of a different vector to it (ii) Change a vector by multiplying it by a non-zero scalar (iii) Remove a zero vector from the set (b) It is easy to see that row reduction operations can change any set of vectors into a basis with pivots.

3 3 (6) The moral is that given a subspace defined as the span of a finite set of vectors, we can find a basis with pivots spanning the same subspace. Since the basis with pivots is a basis (it is linearly independent), it tells us the dimension of the subspace. If we have a (finite) spanning set for a subspace, we can compute its dimension. 4. Equations with pivots (1) A set of homogeneous linear equations is said to be equations with pivots if each equation has a pivot variable with coefficient 1 that appears in no other equation. (2) This is the same as saying that the coefficient vectors for the equations form a basis with pivots. (3) Given any finite set of homogeneous linear equations defining a subspace, if you take the coefficient vectors and row-reduce them to a basis with pivots, you get a set of equations with pivots defining the same subspace. 5. Complementary bases with pivots (1) Suppose you have two bases with pivots in F n, S {v 1,..., v s } and T {w 1,..., w t }. We say that the bases are complementary if s + t n, if they do not share any pivot positions and if v i w j 0 for all 1 i s and 1 j t. (2) In this case every position is a pivot position in exactly one of the sets. (3) Lemma: If two bases with pivots S {v 1,..., v s } and T {w 1,..., w n s } have complementary pivot positions c 1,..., c s and d 1,..., d n s, then for all 1 i s and 1 j n s we have: Proof. v i w j v i w j v i [d j ] + w j [c i ] n v i [k]w j [k] v i [c k ]w j [c k ] + δ ik w j [c k ] + w j [c i ] + v i [d j ] v i [d k ]w j [d k ] v i [d k ]δ jk (4) Given a basis with pivots S, this result tells us how to construct a complementary basis with pivots T. First fill in the complementary pivot positions in T, w i [d j ] δ ij then use the lemma to fill in the remaining positions w i [c j ] v j [d i ] (5) If you think of S as equations and T as a spanning set, then S defines the same subspace at T.

4 4 DAVID MEREDITH Proof. Certainly the vectors in T are in the subspace defined by the equations S. On the other hand, suppose v i x 0 for all v i S. We must show that x span(t ). Let c i be the pivot positions for S and d j be the pivot positions for T. Let u x t i1 x[d i]w i. We will show that u 0, which shows that x span(t ). Since v i x 0, we know for 1 i s: n 0 v i [j]x[j] v i [c j ]x[c j ] + δ ij x[c j ] + x[c i ] + To show u 0, it suffices to show that u[j] 0 for 1 j n. We will show u[c j ] 0 for 1 j s and u[d j ] 0 for 1 j t. u[c i ] x[c i ] x[d j ]w j [c i ] x[d j ]w j [c i ] (v i [d j ] + w j [c i ]) x[d j ] 0 because v i [d j ] + w j [c i ] 0 for all i, j if S and T are complementary u[d i ] x[d i ] x[d i ] x[d j ]w j [d i ] x[d j ]δ ji x[d i ] x[d i ] 0 All our problems are now solved. 6. The questions answered (1) Given a spanning set for a subspace, we can find equations for the subspace. First row reduce to get a basis with pivots, then find the complementary equations.

5 5 (2) Given equations, we can find a spanning set even a basis. Again, row reduce to get a set of equations with pivots, then find the complementary spanning set. Department of Mathematics San Francisco State University San Francisco, CA address: meredith@sfsu.edu URL: