# Subspaces of R n LECTURE Subspaces

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1 LECTURE 7 Subspaces of R n Subspaces Definition 7 A subset W of R n is said to be closed under vector addition if for all u, v W, u + v is also in W If rv is in W for all vectors v W and all scalars r R, then we say that W is closed under scalar multiplication A non-empty subset W of R n that is closed under both vector addition and scalar multiplication is called a subspace of R n Example 72 Let u =(, ) and v =(, 2) be vectors in R 2 We can construct a subset that closed under vector addition as follows W = {w R n w = ju + kv ; j, k positive integers} To see that this set is closed under vector addition, let w,w W Then w = ju + kv w = j u + k v for some positive integers j, k, j, and k But then there are positive integers j, k, j and k such that w + w =(ju + kv) +(j u + k v)=(j + j ) u +(k + k ) v W because both (j + j ) and (k + k ) ar e positive integers if j, k, j, and k are positive integers The set W is not a subspace, however; because it is not closed under scalar multiplication To see this, note that the vector can not be represented as ( ) u 2 = 2, sum of u and v with positive integer coefficients Example 73 The preceding example, however, does provide a clue as to one way to constructing a subspace Let u =(, ) and v =(, 2) be vectors in R 2 Consider the set W = {w R n w = ju + kv ; j, k R} This is closed under vector addition because if w, w W, then there are real numbers r, s, r and s such that But then w = ru + sv w = r u + s v w + w =(r + r ) u +(s + s ) v W since (r + r ) R and (s + s ) R And, for any real number t tw =(tr)u +(ts)v W since (tr) R and (ts) R The following theorem generalizes this last example

2 2 SOLUTIONS OF HOMOGENEOUS SYSTEMS 2 Theorem 74 Let {v,,v k } R n be a set of vectors in R n Then the span of {v,,v k } is a subspace of Proof Recall that the span ofa set ofvector is the set ofall possible linear combinations ofthose vectors Set W = span (v,,v k ) {w R n w = c v + c 2 v c k v k ; c,c 2,,c k R} Then for any vectors w,w W we have w = c v + c 2 v c k v k w = c v + c 2v c kv k for some choice of real numbers c,,c k and c,,c k But then w + w =(c + c )v +(c 2 + c )v (c k + c )v k k W and if t is any real number tw =(tc )v +(tc 2 )v 2 + +(tc k )v k W Remark 75 We shall often refer to the span of a set {v,,v k } ofvectors in R n generated by {v,,v k } as the subspace 2 Solutions of Homogeneous Systems We now come to another fundamental way of realizing a subspace of R n Definition 76 A linear system of the form Ax = is called homogeneous A homogeneous linear system is always solvable since x = is always a solution As such, this solution is not very interesting; we call it the trivial solution A homogeneous linear system may possess other non-trivial solutions (ie solutions x ), this is where we shall focus our attention today Lemma 77 Suppose x and x 2 are solutions of a homogeneous system Ax = Then so is any linear combination rx + sx 2 of x and x 2 Proof Since x and x 2 are solution of Ax = we have Ax = = Ax 2 But then A (rx + sx 2 ) = A(rx )+A(sx 2 ) = r (Ax )+s (Ax 2 ) = r + s = so rx + sx 2 is also a solution Theorem 78 The solution space of a homogeneous linear system is a subspace of R n Proof The preceding lemma demonstrates that the solution space ofa homogeneous linear system is closed under both vector addition (take r = and s = in the proofofthe preceding lemma) and scalar multiplication (let r be any real number and take s =, in the proofofthe lemma) Therefore, it is a subspace ofr n

3 4 BASES 3 3 Subspaces Associated withmatrices Definition 79 The row space of an m n matrix A is the span of row vectors of A Since the row vectors ofan m n matrix are n-dimensional vectors, the row space ofan m n matrix is a subspace of R n Definition 7 The column space of an m n matrix A is the span of column vectors of A Since the column vectors ofan m n matrix are m-dimensional vectors, the column space ofan m n matrix is a subspace of R m Definition 7 The null space of an m n matrix A is the solution set of homogeneous linear system Ax = By the theorem ofthe proceding section, the null space ofan m n matrix A will be a subspace of R n Consider now a non-homogeneous linear system Ax = b The left hand side of such an equation is a a 2 a n a 2 a 22 a 2n am am2 amn x x 2 xn = ax + a2x2 + anxn = x a a 2 a 2 x + a 22 x a 2nxn amx + am2x amnxn am + x 2 a 2 a 22 am2 + x n The final expression on the right hand side is evidently a linear combination ofthe column vectors ofa The consistency ofthe equation Ax = b then requires the column vector b to also lie within the span of the column vectors of A Thus we have Theorem 72 A linear system Ax = b is consistent if and only if b lies in the column space of A a n a 2n amn 4 Bases Consider the subspace generated by the following three vectors in R 3 : v =, v2 =, v3 = It turns out that this is the same as the subspace generated from just v and v 2 To see this note that v 3 = = = v +( )v 2

4 4 BASES 4 But any vector w in span (v, v 2, v 3 ) is expressible in the form w = c v + c 2 v 2 + c 3 v 3 = c v + c 2 v 2 + c 3 (v v 2 ) = (c + c 3 ) v +(c 2 c 3 ) v 2 span (v, v 2 ) For reasons ofeffficiency alone, it is natural to try to find the minimum number ofvectors needed to specify every vector in a subspace W Such a set will be called a basis for W There is also another reason to be interested in basis vectors Consider the vector w = 2 In terms ofthe vectors v, v 2,andv 3, we can write this as either or or even w =3v v 2 2v 3 w = 2v +4v 2 +3v 3 w = 2v +4v 2 +3v 3 However, there is only one way to represent w as a linear combination ofthe vectors v and v 2 For the condition w = c v c 2 v 2 requires 2 = c + c2 is equivalent to the following linear system c + c 2 = 2 c 2 = c = which obviously as c = and c 2 = as its only solution This motivates the following definition = c + c 2 c 2 c Definition 73 Let W be a subspace of R n A subset {w,w 2,,w k } of W is called a basis for W if every vector in W can be uniquely expressed as linear combination of the vectors w,w 2,,w k Theorem 74 A set of vectors {w, w 2,, w k } is a basis for the subspace W generated by {w, w 2,, w k } ifandonlyif r w + r 2 w r k w k = implies =r = r 2 = = r k Proof Suppose {w, w 2,, w k } is a basis for W = span (w, w 2,, w k ) Then every vector in W can be uniquely specfied as a vector ofthe form v = r w + r 2 w r k w k ; r,r 2,,r k R In particular, the zero vector (7) = ()w +()w 2 + +()w k

5 4 BASES 5 lies in W Because {w, w 2,, w k } is assumed to be a basis, the linear combination on the right hand side of(7) must be the unique linear combination ofthe vectors w,,w k that is equal to Hence, r w + r 2 w r k w k = implies = r = r 2 = = r k Suppose r w + r 2 w r k w k = implies = r = r 2 = = r k We want to show that {w, w 2,, w k } is a basis for W = span (w, w 2,, w k ) In other words, we need to show that there is only one choice ofcoefficients r,,r k such that a vector v W can be expressed in the form v =r w + r 2 w r k w k Suppose there were in fact two distinct ways of representing v : (72) (73) v = r w + r 2 w r k w k v = s w + s 2 w s k w k Subtracting the second equation from the first yields =(r s ) w +(r 2 s 2 ) w 2 + +(r k s k ) w k Our hypothesis now implies =r s = r 2 s 2 = = r k s k In other words r = s r 2 = s 2 r k = s k and so the two linear combinations on the right hand sides of(72) and (73) must be identical Theorem 75 Let A be an n n matrix Then the following statements are equivalent The linear system Ax = b has a unique solution for each vector b R n 2 The matrix A is row equivalent to the identity matrix 3 The matrix A is invertible 4 The column vectors of A form a basis for R n Proof We have already demonstrated the equivalence ofstatements 2, 3 and 4 It therefore suffices to show that statement 4 is equivalent to statement To see that statement 4 implies statement, suppose that the column vectors c, c 2,,c n of A form a basis for R n Then by Theorem 72, the linear system Ax = b is consistent for all vectors b span (c,,c n )= R n But a direct calculation reveals b = Ax = xc + x2c xnc n Because the vectors c,,c n form a basis, there choice of coefficients x, x2,,xn must be unique Therefore, the linear system Ax = b has a unique solution for each vector b R n On the other hand, suppose the linear system Ax = b has a unique solution for each vector b R n In particular, this must be true for b = Therefore, there is only one choice of coefficients x,x2,,xn such that = xc + x2c xnc n = Ax By the preceding theorem we can conclude that the column vectors of A form a basis for R n Example 76 Show that the vectors v =(,, 3), v 2 =(3,, 4), and v 3 =(, 4, ) form a basis for R 3

6 By the preceding theorem, it suffices to show that the matrix is invertible Row reducing A yields A = R 2 R 2 R R 3 R 3 3R 4 BASES R3 R R The matrix on the far right is upper triangular so it s obviously invertible Hence, A is invertible; hence the column vectors of A form a basis for R 3 ; hence the vectors v, v 2, and v 3 form a basis for R 3 The preceding theorem is applicable only to square matrices A and linear systems of n equations in n unknowns It can be extended to more general matrices and linear systems in the following manner Theorem 77 Let A be an m n matrix Then the following are equivalent Each consistent system Ax = b has a unique solution 2 The reduced row echelon form of A consists of the n n identity matrix followed by m n rows containing only zeros 3 The column vectors of A form a basis for the column space of A Proof 2 : From Theorem 58 oflecture 5 (Theorem 7 in text), we know that a consistent linear system Ax = b has a unique solution ifand only ifa is row equivalent to a matrix A in row-echelon form such that every column of A has a pivot Since A, and hence A, has n columns, we can conclude that the solution ofevery consistent linear system Ax = b is unique ifand only ifwe have must have n pivots In order to have n pivots the number m ofrows must be n When n = m there will be one pivot for each row, and the pivots will all reside along the diagonal, like so A = a a 2 a n a22 a2n ann occuring in at least n rows If A is further reduced to a matrix A in reduced row-echelon form, then all the pivots are re-scaled to and all the entries above the pivots are equal to Thus, A = = I If m>n, we still require n pivots, that means the only way we can consistently add rows to the picture above is by adding rows without pivots; ie, rows containing only s 3 : Suppose Ax = b has a unique solution for each b in the column space C(A) ofa (Recall b must lie in the column space of A in order for the linear system to be consistent) Then, if we denote the column vectors of A by c, c 2,,c n we have b = Ax x c + x 2 c x n c n, for all b C(A) span (c, c 2,,c n ) Ifthe solution x is unique, then there is only one such linear combination ofthe column vectors c i for each vector b C(A) Hence, the column vectors c i provide a basis for C(A) On the other hand, ifthe column vectors were not a basis for C(A) span (c, c 2,,c n ), then that would mean that there are vectors b

7 4 BASES 7 lying in C(A) such that the expansion b = x c + x 2 c x n c n in terms ofthe c i is not unique Hence, a solution x to Ax = b would not be unique Theorem 78 Let Ax = b be a non-homogeneous linear system, and let p be any particular solution of this sytem Then every solution of Ax = b can be expressed in the form x = p + h where h is a solution of the corresponding homogeneous system Ax = Proof Suppose p and x are both solutions of Ax = b Then set h = x p Then h satisfies Hence, x = p + h with h a solution of Ax = Ah = A (x p) =Ax Ap = b b =

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