Subspaces, basis, dimension, and rank Math 40, Introduction to Linear Algebra Wednesday, February 8, 2012 n Subspaces of R Subspaces of Rn algebraicalgebraic generalization generalization of geometric of OneOne motivation motivation for notionnotion n geometricexamples of linesexamples and planes of lines through and of ofsubspaces subspaces of R n . R planes throughthe origin the origin Consider R5. 0 1 1 1 0 1 1 1 0 0 point 2 2 1 1 2 2 plane 0 t 3 line s 1 + t 3 0 0 4 1 4 (no0 direction t 3 (one 5 direction s 1 +1 t 3 (two5 direction 0 vectors) 4 vector) 1 4 vectors)} point line plane 0 5 1 5 (no direction (one direction (two direction vectorslinearly) independent vectors) vector) linearly independent 1 1 1 1 1 1 1 1 1 1 1 1 2 01 1 11 12 1 r 0 +1 s 1 2+ t 3 p 00 + r 10 + s 11 + t 23 0 1 4 0 0 1 4 r 0 +0s 1 + 1t 3 name?5 p 00 + r 0 + s 11 + t 35 name? 0 1name?4 0 0name? 1 4 0(three direction1 5 vectors ) 0(four direction0 vectors1 )5 linearly independent linearly independent Definition A subspace S of Rn is a set of vectors in Rn such that (1) 0 S [contains zero vector] ∈ (2) if u,v S,thenu + v S [closed under addition] ∈ ∈ (3) if u S and c R,thencu S [closed under scalar multiplication] ∈ ∈ ∈ We can combine (2) and (3) by saying that S is closed under linear combi- nations. Examples a Ex 1: Is S = b : a, b R asubspaceofR3? Yes! • ∈ 0 Lecture 10 Math 40, Spring ’12, Prof. Kindred Page 1 Subspace Definition A subspace S of Rn is a set of vectors in Rn such that (1) 0 S [ contains zero vector ] ∈ (2) if u, v S,thenu + v S [ closed under addition ] ∈ ∈ (3) if u S and c R,thencu S [ closed under scalar mult. ] ∈ ∈ ∈ Subspace Example Definition A subspace S of Rn is a set of vectors in Rn such that (1) 0 S x ∈ Is S = [ y ]:x 0,y 0 (2) if u, v S,thenu + v S { ≥ 2 ≥ } ∈ ∈ a subspace of R ? (3) if u S and c R,thencu S ∈ ∈ ∈ No! 1 1 1 S but = − S 1 ∈ − 1 1 ∈ − = ⇒ S is not closed under scalar multiplication Subspace Example Definition A subspace S of Rn is a set of vectors in Rn such that a (1) 0 S b ∈ Is S = : a, b R (2) if u, v S,thenu + v S 0 ∈ ∈ ∈ 3 a subspace of ? (3) if u S and c R,thencu S R ∈ ∈ ∈ Yes! 0 (1) 0= 0 S contains zero vector 0 ∈ ⇒ a1 a2 (2) Let u, v S. Then u = b1 and v = b2 for some a1,a2,b1,b2 R. ∈ 0 0 ∈ It follows that a1+a2 closed under u + v = b1+b2 S addition 0 ∈ ⇒ a (3) Let u S, c R. Then u = b for some a, b R. It follows that ∈ ∈ 0 ∈ a ca closed under cu = c b = cb S scalar mult. 0 0 ∈ ⇒ Span is a subspace! n Theorem. Let v1, v2,...,vk R . Then S =span(v1, v2,...,vk ) ∈ is a subspace of Rn. Proof. We verify the three properties of the subspace definition. (1) 0=0v +0v + +0v 1 2 ··· k 0 is a linear comb. of v1, v2,...,vk 0 S ⇒ ⇒ ∈ (2) Let u, w S. Then u = c v + + c v and ∈ 1 1 ··· k k w = d v + + d v for some scalars c ,d. Thus, 1 1 ··· k k i i u + w =(c v + + c v )+(d v + + d v ) 1 1 ··· k k 1 1 ··· k k =(c1 + d1)v1 + +(ck + dk )vk u + w S ··· ⇒ ∈ linear comb. of v1,...,vk (3) Let u S, c R. You finish the proof (show that cu S). ∈ ∈ ∈ Column and row spaces of a matrix Definition For an m n matrix A with column vectors m × v1,v2,...,vn R ,thecolumn space of A is span(v1,v2,...,vn). ∈ col(A) is a subspace of Rm since it is the span of a set of vectors in Rm Definition For an m n matrix A with row vectors n × r1,r2,...,rm R ,therow space of A is span(r1,r2,...,rm). ∈ row(A) is a subspace of Rn since it is the span of a set of vectors in Rn Characterizing column and row spaces Important relationships: • Column space scalars x ,x ,...,x such that ∃ 1 2 n b is in Ax = b b col(A) span of | | is consistent ∈ ⇐⇒ ⇐⇒ x1 v1 + xn vn = b cols of A ··· ⇐⇒ (has a sol’n) | | • Row space T T b row(A) b T col(AT ) A x = b has a ∈ ⇐⇒ ∈ ⇐⇒ solution since columns of AT are the rows of A or b is A Ri +kRj A b row(A) linear comb. for i>j ∈ ⇐⇒ of rows of A ⇐⇒ b −−−−→ 0 Null space of a matrix Definition For an m n matrix A,thenull space of A is the set of all × solutions to Ax = 0, i.e., null(A) null(A)= x : Ax =0 . { } null(A) is a set of vectors in Rn Question Is null(A) a subspace of Rn? Yes! This statement requires proof, and we will tackle this on Friday. Basis Definition A set of vectors B = v ,...,v is a basis for a { 1 k } subspace S of Rn if span(B)=S, • and B is a linearly independent set. • 1 0 0 Example Standard basis for 3 is 0 , 1 , 0 R 0 0 1 choice of basis is not 1 0 1 unique but another basis for 3 is 1 , 1 , 0 . R 0 1 1 More on basis 1 0 0 3 Definition A set of vectors B = v1,...,vk is a basis for a Standard basis for R is 0 , 1 , 0 n { } subspace S of R if 0 0 1 span(B)=S, • 1 0 1 but another basis for 3 is 1 , 1 , 0 . and B is a linearly independent set. R • 0 1 1 Do you believe such bases Consider exist for R3? x y 1 1 No! B = x , y 1 2 2 x3 y3 Why not? 3 x1 y1 z1 w1 span(B1) = R • B2 = x2 , y2 , z2 , w2 B2 not linearly indep. x3 y3 z3 w3 • proof by Dimension contradiction Theorem. Any two bases of a subspace have the same number of vectors. Definition The number of vectors in a basis of a subspace S is called the dimension of S. n Example dim(R )=n 1 0 0 0 1 0 n since e1, e2,...,en = . , . ,...,. , is a basis for R { } . 0 0 1 Side-note The trivial subspace 0 has no basis { } since any set containing the zero vector is linearly dependent, so dim( 0 ) = 0. { } Important basis results Theorem. Given a basis B = v ,...,v of subspace S, there is { 1 k } a unique way to express any v S as a linear combination of ∈ basis vectors v1,...,vk . Proof sketch on Friday. n Theorem. The vectors v1,...,vn form a basis of R if and only if { } rank(A)=n,whereA is the matrix with columns v1,...,vn. Proof sketch ( ). Ax = 0 has only cols of A are ⇒ trivial sol’n x = 0 ⇒ linearly indep. ⇒ ⇒ cols of A are rank(A)=n RREF of A basis for n ⇒ is I ⇒ Ax = b is consistent cols of A R n for any b Rn span . ∈ ⇒ R ⇒ Same ideas can be used to prove converse direction. Fundamental Theorem of Invertible Matrices (extended) Theorem. Let A be an n x n matrix. The following statements are equivalent: • A is invertible. • A x = b has a unique solution for all b R n . ∈ • A x = 0 has only the trivial solution x = 0 . • The RREF of A is I. • A is the product of elementary matrices. • rank(A) = n. • Columns of A form a basis for R n. Finding bases for fundamental subspaces of a matrix row(A) Given matrix A, how do we col(A) ? find bases for subspaces {null(A) EROs First, get RREF of A. A R −−−→ Finding bases for fundamental subspaces of a matrix basis for basis for nonzero EROs do not change = = row space of a matrix. row(A) row(R) ⇒ rows of R Columns of A have the columns of A that basis for same dependence = correspond to columns relationship col(A) ⇒ as columns of R. of R w/leading 1’s • solve Ax = 0, i.e. solve Rx = 0 • express sol’ns in terms of free variables, e.g., x1 basis vectors x = x + x 2 1 3 for null(A) x3 Example of matrix subspaces’ bases 12345 10 1 2 3 EROs − − − A = 6 7 8 9 10 A R = 01 2 3 4 −−−→ 11 12 13 14 15 00 0 0 0 basis for = 10 1 2 3 , 01234 row(A) − − − 1 2 basis for = 6 , 7 col(A) 11 12 Example of matrix subspaces’ bases 12345 10 1 2 3 EROs − − − A = 6 7 8 9 10 A R = 01 2 3 4 −−−→ 11 12 13 14 15 00 0 0 0 1 2 3 x1 x3 2x4 3x5 =0 2 3 4 − − − basis for − − − x +2x +3x +4x =0 = 1 , 0 , 0 2 3 4 5 null(A) 0 1 0 x3,x4,x5 free 0 0 1 x1 x3 +2x4 +3x5 1 2 3 x2 2x3 3x4 4x5 2 3 4 − − − − − − x = x3 = x3 = x3 1 + x4 0 + x5 0 x x 0 1 0 4 4 x x 0 0 1 5 5 Example related to column space 101 2 1 Is b col(A)? A = 110 b = 3 c = 1 ∈ Is c col(A)? 000 0 1 ∈ Determine if Ax = b has a solution.