AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda 4. BASES AND DIMENSION

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1 4. BASES AND DIMENSION

2 Definition Let u 1,..., u n be n vectors in V. The vectors u 1,..., u n are linearly independent if the only linear combination of them equal to the zero vector has only zero scalars; that is, given λ 1,..., λ n R if λ 1 u λ n u n = 0 V λ 1 =... = λ n = 0. Otherwise, it is said that u 1,..., u n are linearly dependent, that is λ 1,..., λ n R not all zero such that λ 1 u λ n u n = 0 V. 1. Let us prove that u 1 = (1, 0, 0), u 2 = (0, 3, 0), u 3 = (0, 0, 5) are linearly independent in R 3. If λ 1 u 1 + λ 2 u 2 + λ 3 u 3 = 0 R 3 then (λ 1, 3λ 2, 5λ 3 ) = (0, 0, 0) and thus λ 1 = λ 2 = λ 3 = {(1, 2, 0), (2, 0, 0), (0, 1, 0), (0, 0, 6)} is a set of linearly dependent vectors because (1, 2, 0) (1/2)(2, 0, 0) 2(0, 1, 0) + 0(0, 0, 6) = (0, 0, 0). 3. If 0 V is an element of C, then C is a set of linearly dependent vectors. 4. u, v V are linearly dependent u = αv, for some α R.

3 Theorem Let V be a real vector space. The number of elements of any generating set of V is greater than or equal to the number of elements of any set of linearly independent vectors of V. Definition Let V be a finitely generated real vector space. A subset B = {v 1,..., v n } of V is a basis of V if it verifies, 1. B is a generating set of V, 2. B is a set of linearly independent vectors. Example Let e i be the vector of R n with zeros in every entry except for the ith entry, which equals 1. B = {e 1,..., e n } is a basis of R n called the standard basis. Theorem (Existence of basis) Every set of generating vectors G of a real vector space V which is finitely generated and nonzero contains a basis B of V. Therefore, every real vector space V finitely generated and nonzero has a basis.

4 Example In V = R 2 the set AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda G = {(1, 0), (1, 1), ( 1, 0), (1, 2)} is a generating set of V and contains the basis B = {(1, 0), (1, 1)} of V, which is obtained by removing vectors of G linearly dependent of the remaining vectors of G. In this case ( 1, 0) = ( 1)(1, 0) and (1, 2) = 2(1, 1) (1, 0). Proposition Let V be a finitely generated real vector space and let B = {v 1,..., v n } be a basis of V. Every vector of V has a unique expression as a linear combination of the vectors of B. Definition Let V be a finitely generated real vector space and let B = {v 1,..., v n } be a basis of V. The coordinates of v V are the scalars in the unique list (λ 1,..., λ n ) R n such that v = λ 1 v λ n v n. We will write (λ 1,..., λ n ) B to mean the coordinates of a vector in the basis B.

5 Example In V = R 3 fix the basis B = {v 1 = (1, 0, 0), v 2 = (0, 2, 0), v 3 = (0, 0, 1)}. The coordinates of v = (2, 1, 1) in B are (2, 1/2, 1) B sinse v = 2v 1 + 1/2v 2 1v 3. Remarks 1. Observe that a vector space has infinitely many bases. 2. If V is a finitely generated vector space then every basis has a finite number of vectors. 3. The set R[x] of all the polynomials in x with real coefficients is a real vector space, which is not finitely generated. A basis of R[x] has infinitely many vectors. Dimension Theorem All the bases of a finitely generated real vector space V {0 V } have the same number of vectors. Definition The dimension of a finitely generated real vector space V, is the number of elements of a basis of V. We denote it by dim V. By convention dim{0 V } = 0.

6 Theorem Let V be a finitely generated real vector space. Every linearly independent set of vectors of V is included in a basis of V. Example Let V = R 3. The set of linearly independent vectors I = {(1, 1, 0), (0, 2, 0)} is a subset of the basis {(1, 1, 0), (0, 2, 0), (0, 0, 1)} of R 3, the vector (0, 0, 1) was added to complete I to a basis of V. Definition The rank of a set of vectors C = {u 1,..., u n } in V is the dimension of the subspace they generate: rank(c) = dim C. Given a finite dimensional vector space V, we fix a basis B. Let M be the matrix whose rows are the coordinates of the vectors of C in the basis B. Then, rank(c) = rank(m). Proposition The rank of a matrix M equals the highest number of row vectors of M (equivalently column vectors) which are linearly independent.

7 5. EQUATIONS OF SUBSPACES Let V be a real vector space and let us fix a basis B = {v 1,..., v n } of V. Let U be a vector subspace of V such that 0 < dimu < dimv.

8 CARTESIAN EQUATIONS Proposition There exists a homogeneous system of linear equations AX = 0, with dim V dim U equations, whose solution set equals the set of the coordinates of all the vectors of U in the basis B, that is {(λ 1,..., λ n ) R n λ 1 v λ n v n U} = = {(s 1, s 2,..., s n ) R n AS = 0} where S = s 1 s 2. s n. Every homogeneous system verifying the previous statement is given the name of system of implicit or cartesian equations of U in the basis B. Let us suppose that dim U = m (with 0 < m < n) and let {u 1,..., u m } be a basis of U. To find the implicit equations of U means to look for conditions on the coordinates (x 1, x 2,..., x n ) B of a vector v in B so that v belongs to U.

9 Let us suppose that the coordinates of u i in the basis B are (u i1,..., u in ) B and built the matrix M of size (m + 1) n : x 1 x 2 x n u 11 u 12 u 1n M = u 21 u 22 u 2n.... u m1 u m2 u mn The vector v belongs to U if it is a linear combination of the vectors in {u 1,..., u m }, then rank(m) = m. This means that all the minors of order m + 1 of M are zero. Each minor of order m + 1 provides a homogeneous linear equation. Since dim U = m we can reduce the system to l = n m equations Cartesian equations of U a 11 x 1 + a 12 x a 1n x n = 0 a 21 x 1 + a 22 x a 2n x n = 0. a l1 x 1 + a l2 x a ln x n = 0,

10 taking l minors of order m + 1 containing a nonzero minor of order m previously fixed. Observe that the cartesian equations of a subspace U in the basis B are not unique. PARAMETRIC EQUATIONS Definition The parametric equations of U in the basis B are a parametric solution of the system of cartesian equations of U in the basis B. If dim U = m the parametric equations are x 1 = u 11 α 1 + u 21 α u m1 α m x Parametric equations of U 2 = u 12 α 1 + u 22 α u m2 α m. x n = u 1n α 1 + u 2n α u mn α m, where α 1, α 2,..., α m are the parameters and u ij R.

11 They can be also written as AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda (x 1, x 2,..., x n ) = = (u 11 α 1 + u 21 α u m1 α m,..., u 1n α 1 + u 2n α u mn α m ). This means that a vector v V, with coordinates (x 1, x 2,..., x n ) B, belongs to U if (x 1, x 2,..., x n ) B = = (u 11 α 1 + u 21 α u m1 α m,..., u 1n α 1 + u 2n α u mn α m ) B = = α 1 (u 11,..., u 1n ) B + α 2 (u 21,..., u 2n ) B... + α m (u m1,..., u mn ) B. If we call u i the coordinate vector (u i1,..., u in ) B, i = 1,..., m then is a basis of U. {u 1, u 2,..., u m }

12 6. OPERATIONS WITH SUBSPACES

13 INTERSECTION OF SUBSPACES Let V be a real vector space with basis B. The intersection of vector subspaces is a vector subspace. We explain next how to obtain the equations of the intersection of two subspaces. Let U 1 and U 2 be vector subspaces of V with systems of cartesian equations a 11 x 1 + a 12 x a 1n x n = 0 Equations of U 1. Equations of U 2 U 1 has l 1 equations and U 2 has l 2 equations. a l1 1x 1 + a l1 2x a l1 nx n = 0, b 11 x 1 + b 12 x b 1n x n = 0. b l2 1x 1 + b l2 2x b l2 nx n = 0,

14 The intersection subspace U 1 U 2 contains those vectors of V whose coordinates (x 1, x 2,..., x n ) B verify the system: Remarks ( ) = a 11 x 1 + a 12 x a 1n x n = 0. a l1 1x 1 + a l1 2x a l1 nx n = 0 b 11 x 1 + b 12 x b 1n x n = 0. b l2 1x 1 + b l2 2x b l2 nx n = 0, 1. The system of cartesian equations of U 1 U 2 is obtained removing from ( ) equations that depend linearly on the remaining equations. This can be achieved obtaining the echelon form of ( ). Thus, the number of cartesian equations of U 1 U 2 is l 1 + l A parametric solution of ( ) provides the parametric equations of U 1 U 2 and from them a basis of U 1 U 2 is obtained.

15 SUM OF SUBSPACES Given subspaces U 1 and U 2 of V, in general U 1 U 2 is not a subspace of V. Let us see a counterexample. Example Given the subspaces of V = R 2 U 1 = {(x, 0) R 2 x R} U 2 = {(0, y) R 2 y R}. The set U 1 U 2 is not a subspace of R 2 since u 1 = (1, 0) U 1, u 2 = (0, 1) U 2 but u 1 + u 2 = (1, 1) / U 1 U 2. Definition Let U 1 and U 2 be subspaces of V. We call sum of U 1 and U 2 to the set U 1 + U 2 = {u 1 + u 2 u 1 U 1, u 2 U 2 }.

16 Proposition The set U 1 + U 2 is a vector subspace of V, it is the smallest subspace that contains U 1 U 2, that is U 1 + U 2 = U 1 U 2. To compute U 1 + U 2 it is important to take into account that if B U1 is a basis of U 1 and B U2 is a basis of U 2 then U 1 + U 2 = B U1 B U2, that is, B U1 B U2 is a generating set of U 1 + U 2. We can obtain a basis B U1 +U 2 of U 1 + U 2 from B U1 B U2 by removing vectors, to obtain a linearly independent set, which is also a generating set.

17 DIRECT SUM OF SUBSPACES Definition Let U 1 and U 2 be subspaces of V. The subspace U 1 + U 2 is a direct sum if U 1 U 2 = {0 V }. We write U 1 U 2. Example Let V = R 3 and fix the standard basis. Let us consider vector subspaces U and W given by their cartesian equations, U { x1 + x 2 + x 3 = 0 x 1 + x 3 = 0, W 2x 2 + x 3 = 0. The coordinates (x 1, x 2, x 3 ) of a vector of U W are a solution of the system x 1 + x 2 + x 3 = 0 x 1 + x 3 = 0 2x 2 + x 3 = 0,

18 whose coefficient matrix has rank 3. By Rouche s Theorem the system has only the zero solution (it is a homogeneous system). This shows that U W = {0 R 3} and then U W, that is the sum is a direct sum. Proposition Let B 1 be a basis of U 1 and B 2 a basis of U 2 then: U 1 U 2 = {0 V } B 1 B 2 is a linearly independent set, that is U 1 U 2 B 1 B 2 is a basis of U 1 + U 2. Definition Two subspaces U 1 and U 2 of a real vector space V are complementary if U 1 U 2 = V. U 1 U 2 = V { U1 + U 2 = V U 1 U 2 = {0 V }

19 Remarks AFFINE AND PROJECTIVE GEOMETRY, E. Rosado & S.L. Rueda 1. Every subspace has a complementary subspace. 2. Let U 1 and U 2 be complementary subspaces of V. If B 1 is a basis of U 1 and B 2 is a basis of U 2 then B 1 B 2 is a basis of V. Theorem Dimension Formula or Grassman Formula Let U 1 and U 2 be subspaces of V. Then dim U 1 + dim U 2 = dim(u 1 + U 2 ) + dim(u 1 U 2 ).