Matrix invertibility Rank-Nullity Theorem: For any n-column matrix A, nullity A +rankA = n Corollary: Let A be an R C matrix. Then A is invertible if and only if R = C and the ⇥ | | | | columns of A are linearly independent. Proof: Let F be the field. Define f : FC FR by f (x)=Ax. ! Then A is an invertible matrix if and only if f is an invertible function. The function f is invertible i↵dim Ker f =0anddimFC =dimFR i↵nullity A =0and C = R . | | | | nullity A =0 i↵ dim Null A =0 i↵Null A = 0 { } i↵the only vector x such that Ax = 0 is x = 0 i↵the columns of A are linearly independent. QED Matrix invertibility examples 123 is not square so cannot be invertible. 456 12 is square and its columns are linearly independent so it is invertible. 34 1 1 2 2 1 3 is square but columns not linearly independent so it is not invertible. 2 3 3 1 4 4 5 Transpose of invertible matrix is invertible Theorem: The transpose of an invertible matrix is invertible. a1 A = v v = . AT = a a 2 1 ··· n 3 2 . 3 2 1 ··· n 3 a 6 n 7 4 5 4 5 4 5 Proof: Suppose A is invertible. Then A is square and its columns are linearly independent. Let n be the number of columns. Then rank A = n. Because A is square, it has n rows. By Rank Theorem, rows are linearly independent. Columns of transpose AT are rows of A, so columns of AT are linearly independent. Since AT is square and columns are linearly independent, AT is invertible. QED More matrix invertibility 1 1 Earlier we proved: If A has an inverse A− then AA− is identity matrix Converse: If BA is identity matrix then A and B are inverses? Not always true. Theorem: Suppose A and B are square matrices such that BA is an identity matrix 1.ThenA and B are inverses of each other. Proof: To show that A is invertible, need to show its columns are linearly independent. Let u be any vector such that Au = 0. Then B(Au)=B0 = 0. On the other hand, (BA)u = 1u = u, so u = 0. 1 1 1 This shows A has an inverse A− . Now must show B = A− . We know AA− = 1. BA = 1 1 1 1 (BA)A− = 1A− by multiplying on the right by A− 1 1 (BA)A− = A− 1 1 B(AA− ) = A− by associativity of matrix-matrix mult 1 B 1 = A− 1 B = A− QED Representations of vector spaces Two important ways to represent a vector space: As the solution set of homogeneous linear system As Span b ,...,b { 1 k } a1 x =0,...,am x =0 · · Equivalently, a1 Equivalently, Null . 2 . 3 Col 2 b b 3 1 ··· k am 6 7 6 7 6 7 4 5 6 7 6 7 4 5 Conversions between the two representations [x, y, z]: { [4, 1, 1] [x, y, z]=0, Span [1, 2, 2] − · { − } [0, 1, 1] [x, y, z]=0 · } Span [4, 1, 1], [0, 1, 1] [x, y, z]:[1, 2, 2] [x, y, z]=0 { − } { − · } Conversions for affine spaces? I From representation as solution set of linear system to representation as affine hull I From representation as affine hull to representation as solution set of linear system Conversions for affine spaces? From representation as solution set of linear system to representation as affine hull I input: linear system Ax = b I output: vectors whose affine hull is the solution set of the linear system. x 121 1 y = 122 2 3 2 − z 1 Let u be one solution to the linear system. u =[ 0.5, 0.754, 0] 5 − x 121 0 2 Consider the corresponding homogeneous system Ax = 0. y = 122 2 3 0 − z Its solution set, the null space of A,isavectorspace . 4 5 V 3 Let b ,...,b be generators for . b =[2, 3, 4] 1 k V 1 − 4 Then the solution set of the original linear system is the affine hull of u, b + u, b + u,...,b + u. [ 0.5, .75, 0] and 1 2 k − − [ 0.5, .75, 0] + [2, 3, 4] − − − From representation as solution set to representation as affine hull x 121 1 One solution to equation y = is u =[ 0.5, 0.75, 0] 122 2 3 2 − − z 121 4 5 Solution set of equation is u +Span b1 , Null space of is Span b1 : { } 122 { } i.e. the affine hull of u and u + b − 1 b 1 u+b1 u How to transform between these two representations? Problem 1 (From left to right): I input: homogeneous linear system a x =0,...,a x = 0, 1 · m · I output: basis b1,...,bk for solution set Problem 2 (From right to left): I input: independent vectors b1,...,bk , I output: homogeneous linear system a x =0,...,a x = 0 whose solution set equals 1 · m · Span b ,...,b { 1 k } Representations of vector spaces Two important ways to represent a vector space: As the solution set of homogeneous linear system As Span b ,...,b { 1 k } a1 x =0,...,am x =0 · · Equivalently, a1 Equivalently, Null . 2 . 3 Col 2 b b 3 1 ··· k am 6 7 6 7 6 7 4 5 6 7 6 7 4 5 Representations of vector spaces Two important ways to represent a vector space: As the solution set of homogeneous linear system As Span b ,...,b { 1 k } a1 x =0,...,am x =0 · · Equivalently, a1 Equivalently, Null . 2 . 3 Col 2 b b 3 1 ··· k am 6 7 6 7 6 7 4 5 6 7 6 7 How to transform between these two representations? 4 5 Problem 1 (From left to right): I input: homogeneous linear system a x =0,...,a x = 0, 1 · m · I output: basis b1,...,bk for solution set Problem 2 (From right to left): I input: independent vectors b1,...,bk , I output: homogeneous linear system a x =0,...,a x = 0 whose solution set equals 1 · m · Span b ,...,b { 1 k } with independent rows with independent columns Reformulating Problem 1 I input: homogeneous linear system a x =0,...,a x = 0, 1 · m · I output: basis b1,...,bk for solution set Let’s express this in the language of matrices: a1 . I input: matrix A = 2 . 3 a 6 m 7 4 5 I output: matrix B = b b 2 1 ··· k 3 such that Col B =NullA 4 5 Can require the rows of the input matrix A to be linearly independent. (Discarding a superfluous row does not change the null space of A.) with independent rows Reformulating Problem 1 I input: homogeneous linear system a x =0,...,a x = 0, 1 · m · I output: basis b1,...,bk for solution set Let’s express this in the language of matrices: a1 . I input: matrix A = 2 . 3 a 6 m 7 4 5 with independent columns I output: matrix B = b b 2 1 ··· k 3 such that Col B =NullA 4 5 Can require the rows of the input matrix A to be linearly independent. (Discarding a superfluous row does not change the null space of A.) Reformulating Problem 1 I input: homogeneous linear system a x =0,...,a x = 0, 1 · m · I output: basis b1,...,bk for solution set Let’s express this in the language of matrices: a1 . I input: matrix A = 2 . 3 with independent rows a 6 m 7 4 5 with independent columns I output: matrix B = b b 2 1 ··· k 3 such that Col B =NullA 4 5 Can require the rows of the input matrix A to be linearly independent. (Discarding a superfluous row does not change the null space of A.) same as requiring number of columns of B = n m − 0 0 ··· = same as requiring AB = . ) 2 . 3 0 0 6 ··· 7 = same as requiring number of columns4 of B =nullity5 A ) I input: m n matrix A with independent rows ⇥ I output: matrix B with n m independent columns such that AB = − 0 h i Reformulating the reformulation of Problem 1 I input: matrix A with independent rows I output: matrix B with independent columns such that Col B =NullA By Rank-Nullity Theorem, rank A +nullityA = n Because rows of A are linearly independent, rank A = m, so m +nullityA = n Requiring Col B =NullA is the same as requiring (i) Col B is a subspace of Null A (ii) dim Col B =nullityA same as requiring number of columns of B = n m − = same as requiring number of columns of B =nullityA ) I input: m n matrix A with independent rows ⇥ I output: matrix B with n m independent columns such that AB = − 0 h i Reformulating the reformulation of Problem 1 I input: matrix A with independent rows I output: matrix B with independent columns such that Col B =NullA By Rank-Nullity Theorem, rank A +nullityA = n Because rows of A are linearly independent, rank A = m, so m +nullityA = n Requiring Col B =NullA is the same as requiring 0 0 ··· (i) Col B is a subspace of Null A = same as requiring AB = . ) 2 . 3 0 0 6 ··· 7 (ii) dim Col B =nullityA 4 5 I input: m n matrix A with independent rows ⇥ I output: matrix B with n m independent columns such that AB = − 0 h i Reformulating the reformulation of Problem 1 I input: matrix A with independent rows I output: matrix B with independent columns such that Col B =NullA By Rank-Nullity Theorem, rank A +nullityA = n Because rows of A are linearly independent, rank A = m, so m +nullityA = n Requiring Col B =NullA is the same as requiring 0 0 ··· (i) Col B is a subspace of Null A = same as requiring AB = .