Generating Sets and Bases

Transcription

1 Lecture 6 Generating Sets and Bases Let V be the vector space R 2 and consider the vectors (, 0), (0, ). Then, every vector (x, y) R 2 can be written as a combination of those vectors. That is: (x, y) = x(, 0) + y(0, ). Similarly, the two vectors (, ) and (, 2) do not belong to the same line, and every vector in R 2 can be written as a combination of those two vectors. 6. Introduction In particular: gives us two equations (x, y) = a(, ) + (, 2) a + b = x and a + 2b = y Thus, by substituting the first equation to the second, we get b = x + y Inserting this into the first equation we get a = 2x y Take for example the point (4, ). Then: (4, ) = 5(, ) + ( )(, 2) = 5(, ) (, 2) 5

2 6 LECTURE 6. GENERATING SETS AND BASES We have similar situation for R and all of the spaces R n. In the case of R, for example, every vector can be written as combinations of (, 0, 0), (0,, 0) and (0, 0, ), i.e., (x, y, z) = x(, 0, 0) + y(0,, 0) + z(0, 0, ). Or, as a combination of (,, 0), (,, ) and (0,, ), that is: (x, y, z) = a(,, 0) + b(,, ) + c(0,, ). The latter gives three equation: a + b = x () a + b + c = y (2) b c = z (). (2) + () gives: (4) + () gives: Then () gives: Finally, () gives: Hence, we get: a + 2b = y + z (4) b = x + y + zorb = x + y + z. a = x b = x x + y + z = 2x y z. c = b z = x + y 2z (x, y, z) = 2x y z (,, 0) + x + y + z (,, ) + x + y 2z (0,, ).

3 6.2. IN GENERAL In general Notice that we get only one solution, so there is only one way that we can write a vector in R as a combination of those vectors. In general, if we have k vectors in R, then the equation: x = (x, x 2,..., x n ) = c v + c 2 v c k v k ( ) gives n-equations involving the n-coordinates of v, v 2,..., v k and the unknowns c, c 2,..., c k. There are three possibilities: 6. Possibilities A The equation(*) has no solution. Thus, x can not be written as a combination of the vectors v, v 2,..., v k. B The equation (*) has only one solution, so x can be written in exactly one way as a combination of v, v 2,..., v k. C The system of equations has infinitely many solutions, so there are more than one way to write x as a combination of v, v 2,..., v k. 6.. Case C Let us look at the last case a little closer. If we write x in two different ways: Then, by subtracting, we get: x = c v + c 2 v c k v k x = d v + d 2 v d k v k 0 = (c d )v + (c 2 d 2 )v (c k d k )v k where some of the numbers c i d i are non-zero. Similarly, since we can write: and then we also have: 0 = a v + a 2 v a k v k x = c v + c 2 v c k v k x = (c + a )v + (c 2 + a 2 )v (c k + a k )v k. Thus, we can write x as a combination of the vectors v, v 2,..., v k in several different ways (in fact -many ways).

4 8 LECTURE 6. GENERATING SETS AND BASES 6.4 Definitions We will now use this as a motivation for the following definitions. Definition. Let V be a vector space and v, v 2,..., v n V.. Let W V be a subspace. We say that W is spanned by the vectors v, v 2,..., v n if every vector in W can be written as a linear combination of v, v 2,..., v n. Thus, if w W, then there exist numbers c, c 2,..., c n R such that w = c v + c 2 v c n v n. 2. The set of vectors v, v 2,..., v n is linearly dependent if there exist c, c 2,..., c n, not all equal to zero, such that c v + c 2 v c n v n = 0. Definition.. The set of vectors v, v 2,..., v n is linearly independent if the set is not linearly dependent(if and only if we can only write c v + c 2 v c n v n = 0 with all c i = 0). 2. The set of vectors v, v 2,..., v n is a basis for W, if v, v 2,..., v n is linearly independent and spans W. Before we show some examples, let us make the following observations: Lemma. Let V be a vector space with an inner product (.,.). Assume that v, v 2,..., v n is an orthogonal subset of vectors in V (thus (v i, v j ) = 0 if i j). If v = c v + c 2 v c n v n, then c i = (v,v i) v i 2, i =,..., n. Proof. Assume that v = c v + c 2 v c n v n. Take the inner product with v in both sides of the equation. The LHS is (v, v ). The RHS is: (c v + c 2 v c n v n, v ) = c (v, v ) + c 2 (v 2, v ) c n (v n, v ) = c (v, v ) = c v 2. Thus, (v, v ) = c v 2, or c = (v,v ) v 2. Repeat this for v 2,..., v n. Corollary. If the vectors v, v 2,..., v n are orthogonal, then they are linearly independent.

5 6.5. EXAMPLES Examples Example. Let V = R 2. because: The vectors (, 2) and ( 2, 4) are linearly dependent ( 2)(, 2) + ( 2, 4) = 0. The vectors (, 2), (, ) are linearly independent. In fact, (, 2), (, ) is a basis for R 2. Thus, Indeed, let (x, y) R 2. Then, (x, y) = c (, 2) + c 2 (, ) = (c + c 2, 2c + c 2 ). x = c + c 2 y = 2c + c 2. Subtracting we get: x y = c, or c = y x. Plugging this into the first equation we get: c 2 = x c = x (y x) = 2x y. Thus, we can write any vector in R 2 as a combination of those two. In particular, for (0, 0) we get c = c 2 = 0. The vectors (, 2), ( 2, ) are orthogonal and hence linearly independent, and in fact a basis. Hence, (x, y) = c (, 2) + c 2 ( 2, ). Taking the inner product we get: c = x+2y v 2 = x+2y 5 and c 2 = 2x+y 5. Example. Let V = R. One vector can only generate a line, two vectors can at most span a plane, so we need at least three vectors to span R. The vectors (, 2, ), (,, ) are orthogonal but not a basis. In fact, those two vectors span the plane: W = (x, y, z) R : x z = 0 (explain why). On the other hand, the vectors: (, 2, ), (,, ) and (, 0, ) are orthogonal, and hence a basis.

6 40 LECTURE 6. GENERATING SETS AND BASES with We have, for example: In general, we have: (4,, ) = c (, 2, ) + c 2 (,, ) + c (, 0, ) (x, y, z) = x + 2y + z 6 c = c 2 = 4 + c = 4 2 (, 2, ) + x y + z Let us now discuss some spaces of functions: (,, ) + x z (, 0, ) 2 a) Let v 0 (x) =, v (x) = x and v 2 (x) = x 2. Then, v o, v and v 2 are linearly independent. Take x = 0, then we get c 0 = 0 Differentiate both sides to get: 0 = c 0 v 0 (x) + c v (x) + c 2 v 2 (x) for all x = c 0 + c x + c 2 x 2 0 = c + 2c 2 x Take again x = 0 to find c = 0. Differentiate one more time to get that c 2 = 0. Notice that the span of v 0, v, v 2 is in the space of polynomials of degree 2. Hence, the functions, x, x 2 form a basis for this space. Notice that the functions + x, 2x, x 2 are also a basis. b) Are the functions v 0 (x) = x, v (x) = xe x linearly independent/dependent on R? Answer: No. Assume that 0 = c 0 x + c xe x. It does not help to put x = 0 now, but let us first differentiate both sides and get: 0 = c 0 + c e x + c xe x Now, x = 0 gives: 0 = c 0 + c ()

7 6.5. EXAMPLES 4 Differentiating again, we get: 0 = c e x + c e x + c xe x. Now, x = 0 gives 0 = 2c, or c = 0. Hence, () gives c = 0. c) The functions χ [0, ), χ [,) are orthogonal and hence linearly independent. 2 2 Let us show this directly. Assume that 0 = c χ [0, 2 ) + c 2χ [ 2,). An equation like this means that every x we put into the function on the RHS, the result is always 0. Let us take x = 4. Then: χ [0, )( 2 4 ) =, but χ [,)( 2 4 ) = 0. Hence, 0 = c + c 2 0, or c = 0. Taking x = 4 shows that c 2 = 0. d) The functions χ [0, ), χ [0,) are not orthogonal, but linearly independent. 2 0 = c χ [0,) + c 2 χ [0, 2 ). Take x so that χ [0, )(x) = 0 but χ [0,)(x) =. Thus, any x [0, ) \ [0, 2 2 ) = [ 2, ) will do the job. So, take x = 4. Then, we see that: Then take x = 4 to see that c 2 = 0. 0 = c + c 2 0, orc = 0.