MAT 242 CHAPTER 4: SUBSPACES OF R n

MAT 242 CHAPTER 4: SUBSPACES OF R n JOHN QUIGG 1. Subspaces Recall that R n is the set of n 1 matrices, also called vectors, and satisfies the following properties: x + y = y + x x + (y + z) = (x + y) + z x + 0 = x x x = 0 c(x + y) = cx + cy (c + d)x = cx + dx (cd)x = c(dx) 1x = x where we used our normal notation for vectors rather than general matrices (and c and d denote real numbers, also called scalars). A subspace of R n is a nonempty subset V of R n which is closed under addition and scalar multiplication, that is: (i) x + y V for all x, y V ; (ii) cx V for all c R and x V. Every subspace contains the zero vector 0, and this is usually how a proposed subspace is verified to be nonempty. Actually, the nonemptiness is not an issue in practice, so we typically won t bother checking it explicitly. Example: The set V := {(x, y) R 2 x = 2y} is a subspace, because if (x, y), (z, w) V and c R then and (x, y) + (z, w) = (x + z, y + w) x + z = 2y + 2w = 2(y + w), Date: October 3, 2004. 1

2 JOHN QUIGG so (x, y) + (z, w) V, and and so c(x, y) V. c(x, y) = (cx, cy) cx = c(2y) = 2(cy), The smallest subspace of R n is {0}, and the biggest is R n itself. Every other subspace is proper. In R 2, every line through the origin is a proper subspace. In R 3, every line or plane through the origin is a proper subspace. The solution set of a linear system is a subspace if and only if the system is homogeneous, in which case the subspace is also called the solution space of the homogeneous system. If A is an m n matrix, the solution space of the homogeneous system Ax = 0 is also called the null space of A and denoted Null A. Example: The subspace V of R 2 in the preceding example is the null space of the matrix [ 1 2 ]. The span of a finite subset {v 1, v 2,..., v k } of R n is { k } span{v 1, v 2,..., v k } := c i v i c 1,..., c k R, that is, the set of all linear combinations of v 1,..., v k. It is a subspace. We also say v 1,..., v k span V if V = span{v 1,..., v k }. Example: The subspace V of R 2 in the first example is the span of the vector (2, 1). The column space of an m n matrix A is the subspace of R m spanned by the columns of A, denoted Col A. Thus, a system Ax = b is consistent if and only if b Col A. To put it another way, i=1 Col A = {Ax x R n } Example: The subspace V of R 2 in the first example is the column space of the matrix [ 2 1 ]. Example: The matrices 2 1 1 1 1 1 1 1 1 0 0 1 and 1 1 1 1 0 1 1 1 0 0 0 1

CHAPTER 4 3 are row equivalent, so (1, 1, 1) is not in the column space of the matrix 1 1 1 2 1 1 1 0 0 The columns of an m n matrix A span R m if and only if the reduced echelon form of A has no zero rows. Example: The columns e 1,..., e n of the n n identity matrix (so that e i = (0,..., 1,..., 0) with 1 in the ith coordinate) span R n. It is convenient to regard the empty set of vectors as spanning the zero subspace {0} of every R n.

4 JOHN QUIGG 2. Linear independence A finite subset {v 1, v 2,..., v k } of R n is linearly dependent if there exist scalars c 1,..., c k such that at least one c i is nonzero and k i=1 c iv i = 0. The vectors are linearly independent if they are not linearly dependent. Example: The vectors (1, 2, 1), (1, 1, 1), (1, 1, 1), and (1, 2, 0) are dependent because 2(1, 2, 1) 3(1, 1, 1) + (1, 1, 1) + 0(1, 2, 0) = (0, 0, 0) Example: (1, 1, 1), (0, 1, 1), and (0, 0, 1) are independent because if a(1, 1, 1) + b(0, 1, 1) + c(0, 0, 1) = (0, 0, 0) then (a, a + b, a + b + c) = (0, 0, 0), from which we deduce in succession a = 0, then b = 0, and finally c = 0. The columns of a matrix A are independent if and only if Ax = 0 has only the trivial solution, if and only if every consistent system Ax = b has a unique solution. If k 2, then v 1,..., v k are dependent if and only if there exists j 2 such that v j is a linear combination of v 1,..., v j 1, if and only if one of the v i s is a linear combination of the others. 2 vectors are dependent if and only if they are parallel, that is, one of them is a scalar multiple of the other. 3 vectors in R 3 are dependent if and only if they are coplanar, that is, lie in a plane. Any finite set of vectors in R n containing 0 is dependent. Any subset of an independent set is also independent. If {v 1,..., v k } is independent and u / span{v 1,..., v k }, then {v 1,..., v k, u} is independent. If {v 1,..., v k } is independent and k i=1 c iv i = k i=1 d iv i, then c i = d i for all i = 1,..., k. If A and B are row equivalent, then a set of columns of A is independent if and only if the corresponding columns of B are independent. Example: The columns e 1,..., e n of the n n identity matrix are independent. If V = span{v 1,..., v n }, w 1,..., w k V, and k > n, then {w 1,..., w k } is dependent.

v 1,..., v k R n are dependent if k > n. CHAPTER 4 5 It is convenient to regard the empty set of vectors as independent.

6 JOHN QUIGG 3. Bases A basis of a subspace V of R n is an independent spanning set for V. Example: The columns e 1,..., e n of the n n identity matrix comprise the standard basis of R n. Example: The empty set is a basis for the subspace {0}. Any two bases of V have the same number of vectors, and this number is the dimension of V, denoted dim V. Example: dim R n = n Example: dim{0} = 0 The leading columns of a reduced echelon matrix A form a basis of Col A. Every spanning set for V contains a basis of V. Example: Let V = span{(1, 2, 0, 1), (2, 4, 0, 2), (2, 1, 1, 2), (5, 7, 1, 5), (0, 0, 1, 0)} Then V = Col A, where A = The reduced echelon form of A is 1 2 2 5 0 2 4 1 7 0 0 0 1 1 1 1 2 2 5 0 1 2 0 3 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 Thus columns 1, 3, and 5 of A form a basis of Col A, so {(1, 2, 0, 1), (2, 1, 1, 2), (0, 0, 1, 0)} is a basis of V contained in the given spanning set. Every independent set in V is contained in a basis of V. Example: The vectors (1, 2, 1, 1) and (2, 1, 3, 0) are independent. The columns of the matrix 1 2 1 0 0 0 A = 2 1 0 1 0 0 1 3 0 0 1 0 1 0 0 0 0 1

CHAPTER 4 7 span R 4 since the last 4 columns do. The reduced echelon form of A is 1 0 0 0 0 1 0 1 0 0 1/3 1/3 0 0 1 0 2/3 1/3 0 0 0 1 1/3 5/3 Thus the 1st 4 columns of A give a basis for Col A, hence {(1, 2, 1, 1), (2, 1, 3, 0), (1, 0, 0, 0), (0, 1, 0, 0)} is a basis of R 4 containing the given independent set. If dim V = n, then vectors v 1,..., v n in V are independent if and only if they span V. If V is a proper subspace of R n (that is, V is different from both {0} and R n ), then 0 < dim V < n. Our standard method of solving a homogeneous system gives a basis for the solution space. Example: Let A = 1 2 0 2 0 0 0 1 0 0 0 0 0 0 1 The augmented matrix of the associated homogeneous system Ax = 0 is 1 2 0 2 0 0 0 0 1 0 0 0 0 0 0 0 1 0 which is already in reduced echelon form. The solution space is ( 2s + 2t, s, 0, t, 0) = s( 2, 1, 0, 0, 0) + t(2, 0, 0, 1, 0) By construction the set {( 2, 1, 0, 0, 0), (2, 0, 0, 1, 0)} spans Null A. The special nature of our standard method ensures that this set is automatically independent, hence is a basis for the solution space Null A.

8 JOHN QUIGG 4. Rank and nullity The rank of a matrix A is rank A := dim Col A. The nullity of a matrix A is nullity A := dim Null A. If A is m n, then rank A + nullity A = n The transpose of an m n matrix A = [a ij ] is the n m matrix A T = [a ji ] obtained by interchanging the rows and columns of A. Example: Properties: [ 1 2 3 4 5 6 (A T ) T = A (A + B) T = (A T + B T ) (ca) T = ca T (AB) T = B T A T ] T = 1 4 2 5 3 6 The row vectors of A are the column vectors of A T. The row space of A is Row A := Col A T. The nonzero row vectors of a reduced echelon matrix A form a basis for Row A. If A and B are row equivalent then Row A = Row B. Example: Let A = The reduced echelon form of A is Thus is a basis for Row A. 1 1 0 1 2 2 0 2 2 3 1 2 0 1 1 0 1 0 1 1 0 1 1 0 0 0 0 0 0 0 0 0 {(1, 0, 1, 1), (0, 1, 1, 0)}

CHAPTER 4 9 rank A = dim Row A, because in a reduced echelon matrix the number of nonzero rows equals the number of leading columns.