1 Last time: inverses

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1 MATH Linear algebra (Fall 8) Lecture 8 Last time: inverses The following all mean the same thing for a function f : X Y : f is invertible f is one-to-one and onto 3 For each b Y there is exactly one a X with f(a) = b 4 There is a unique function f : Y X, called the inverse of f, such that f (f(a)) = a and f(f (b)) = b for all a X and b Y Proposition If T : R n R m is linear and invertible then m = n and T is invertible The following all mean the same thing for an n n matrix A: A is invertible A is the standard matrix of an invertible linear function T : R n R n 3 There is a unique n n matrix A, called the inverse of A, such that A A = AA = I n where we define I n = 4 For each b R n the equation Ax = b has a unique solution 5 RREF(A) = I n 6 The columns of A are linearly independent and their span is R n [ a b Proposition Let A = be a matrix c d () A is invertible if and only if ad bc [ () If ad bc then A = d b ad bc c a Proposition Let A and B be n n matrices If A is invertible then (A ) = A If A and B are both invertible then AB is invertible and (AB) = B A 3 If A is invertible then A T is invertible and (A T ) = (A ) T Process to compute A Let A be an n n matrix Consider the n n matrix [ A I n If A is invertible then RREF ([ A I n ) [ = In A So to compute A, row reduce [ A I n to reduced echelon form, and then take the last n columns

2 MATH Linear algebra (Fall 8) Lecture 8 Stronger characterization of invertible matrices Remember that a matrix can only be invertible if it has the same number of rows and columns Theorem When A is an n n matrix, the following are equivalent: (a) A is invertible (b) The columns of A are linearly independent (c) The span of the columns of A is R n Proof We already know that (a) implies both (b) and (c) Assume just (b) holds Then A has a pivot position in every column, so RREF(A) = I n since A has the same number of rows and columns But this implies that A is invertible Similarly, if (c) holds then A has a pivot position in every row, so RREF(A) = I n and A is invertible Corollary Suppose A and B are n n matrices If AB = I n then BA = I n This means that if we want to show that B = A then it is enough to just check that AB = I n Proof Assume AB = I n Then the columns of A span R n since if v R n then Au = v for u = Bv R n, so A is invertible Therefore B = A AB = A I n = A so BA = A A = I n 3 Subspaces of R n Let n be a positive integer Write = Rn Definition Let H be a subset of R n The subset H is a subspace if these three conditions hold: H u + v H for all u, v H 3 cv H for all c R and v H Common examples R n is a subspace of itself The set {} consisting of just the zero vector is a subspace of R n The empty set is not a subspace since it does not contain A subset H R is a subspace if and only if H = {} or H = R or H is the set of all scalar multiples of a single nonzero vector The span of any set of vectors in R n is a subspace Later, we will see that every subspace is the span of some set of vectors

3 MATH Linear algebra (Fall 8) Lecture 8 Example The set is not a subspace since / X Example The set X = v = v v R 3 : v + v + v 3 = v 3 H = v = is a subspace since if u, v H and c R then v v v 3 R 3 : v + v + v 3 = (u + v ) + (u + v ) + (u 3 + v 3 ) = (u + u + u 3 ) + (v + v + v 3 ) = + = and so u + v H and cv H cv + cv + cv 3 = c(v + v + v 3 ) = Any matrix A gives rise to two subspaces, called the column space and null space Definition The column space of an m n matrix A is the subspace Col A R m given by the span of the columns of A Remark If T : R n R m is the linear function T (x) = Ax then Col A = range(t ) Note that Col A = R m if and only if Ax = b has a solution for each b R m A vector b R m belongs to Col A if and and only if Ax = b has a solution Definition The null space of an m n matrix A is the subspace given by the set of vectors v R n with Av = Nul A R n Proof that Nul A is a subspace If u, v Nul A and c R then A(u + v) = Au + Av = + = and A(cv) = c(av) =, so u + v Nul A and cv Nul A Thus Nul A is a subspace of R n Remark If T : R n R m is the linear function T (x) = Ax then Nul A = {x R n : T (x) = } The column space is a subspace of R m where m is the number of rows of A, while the null space is a subspace of R n where n is the number of columns of A Each subspace is completely determined by a finite amount of data This data will be called a basis Definition Let H be a subspace of R n A basis for H is a set of vectors {v, v,, v k } H that are linearly independent and have span equal to H The empty set = {} is considered to be a basis for the zero vector space {} 3

4 MATH Linear algebra (Fall 8) Lecture 8 Example The vectors {e, e,, e n } R n where e = We call this the standard basis of R n Theorem Every subspace H of R n has a basis of size at most n Proof If H = {} then is a basis, e =, an so on, is a basis for Rn Assume H {} Let B be a set of linearly independent vectors in H that is as large as possible The size of B must be at most n since any n + vectors in R n are linearly dependent Let w, w,, w k be the elements of B Since B is as large as possible, if v H is any vector then w, w,, w k, v are linearly dependent so we can write c w + c w + + c k w k + cv = for some numbers c, c,, c k, c R which are not all zero If c = then this would imply that the vectors in B are linearly dependent But the vectors in B are linearly independent, so we must have c Therefore v = c c w + c c w + + c k c w k This means that v is in the span of the vectors in B Since v H is an arbitrary vector, we conclude that the span of the vectors in B is all of H, so B is a basis for H Example Let A = How can we find a basis for Nul A? Well, finding a basis for Nul A is more or less the same task as finding all solutions to the homogeneous equation Ax = So let s first try to solve that equation If we row reduce the 3 6 matrix [ A, we get [ 3 A = RREF( [ A ) { x x x 4 + 3x 5 = This tells us that Ax = if and only if x 3 + x 4 x 5 = Therefore x Nul A if and only if x x + x 4 3x 5 x x = x 3 x 4 = x x 4 + x 5 x 4 x 5 x 5 The vectors, = x, + x x 5 3 4

5 MATH Linear algebra (Fall 8) Lecture 8 are a basis for Nul A: we just computed that these vectors span the null space, and they are linearly independent since each has a nonzero entry in a row (namely, either row, 4, or 5) where the others have zeros (Why does this imply linear independence?) This example is important: the procedure just described works to construct a basis of Nul A for any matrix A The size of this basis will always be equal to the number of free variables in the linear system Ax = How to find a basis for Nul A is something you should remember at the end of this course Example Let B = 3 5 This matrix is in reduced echelon form Finding a basis for Col B is in some ways easier than finding a basis for Nul B The columns of B automatically span Col B, but they might not be linearly independent The largest linearly independent subset of the columns of B will be a basis for Col B, however In our example, the pivot columns, and 5 are linearly independent since each has a row with a where the others have s These columns span columns 3 and 4, so it follows that,, is a basis for Col B This example was special since the matrix B was already in reduced echelon form To find a basis of the column space of an arbitrary matrix, we rely on the following observation: Proposition Let A be any matrix The pivot columns of A form a basis for Col A Proof This proof sketches the main ideas but doesn t spell out all the details Suppose A is m n The reduced echelon form of A is obtained by multiplying A by an invertible matrix E on the left, so we can write RREF(A) = EA If a, a,, a k are the pivot columns of A, then E [ [ Ik a a a k is the m k matrix where the means an (m k) n submatrix of zeros These columns are linearly independent since if [ a a a k v = for v R k then which implies that v = = E [ a a a k v = [ Ik [ v v = Suppose w is a non-pivot column of A The definition of reduced echelon form implies that the corresponding column Ew of RREF(A) = EA is in the span of Ea, Ea,, Ea k (Why?) If we have Ew = c Ea + + c k Ea k then multiplying both sides by E gives w = c a + + c k a k so w is in the span a, a,, a k Therefore the span of the pivot columns of A contains all of the other columns, so is equal to Col A Since the pivot columns are linearly independent and have span equal to Col A, they form a basis 5

6 MATH Linear algebra (Fall 8) Lecture 8 Example The matrix A = is row equivalent to the matrix B in the previous example The pivot columns of A are therefore also columns,, and 5, so 3 9, 3, is a basis for Col A Next time: we will show that if H is a subspace of R n then all of its bases have the same size The common size of each basis is the dimension of H 6

7 MATH Linear algebra (Fall 8) Lecture 8 4 Vocabulary Keywords from today s lecture: Subspace of R n A subset H R n such that H; if u, v H then u + v H; and if v H, c R then cv H Example: Pick any vectors v, v,, v p R n Then R-span{v, v,, v p } is a subspace Column space of an m n matrix A The subspace Col A = {Av : v R n } R m The span of the columns of A x Example: If A = then Col A = y R 3 : x, y R 3 Null space of an m n matrix A The subspace Nul A = {v R n : Av = } R n [ Example: If A = then Nul A = 4 Basis of a subspace H R n x x y R 3 : x, y R = R-span, A set of linearly independent vectors in H whose span is H Example: The vectors, are a basis for the subspace The standard basis of R n consists of the vectors e =, e = v v v 3 R 3 : v + v + v 3 =,, e n = 7