Matrix invertibility Rank-Nullity Theorem: For any n -column matrix A - PowerPoint PPT Presentation

Matrix invertibility Rank-Nullity Theorem: For any n -column matrix A , nullity A + rank A = 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. → F R by f ( x ) = A x . Proof: Let F be the field. Define f : F C − Then A is an invertible matrix if and only if f is an invertible function. dim Ker f = 0 and dim F C = dim F R The function f is invertible i ff i ff nullity A = 0 and | C | = | R | . nullity A = 0 i ff dim Null A = 0 Null A = { 0 } i ff the only vector x such that A x = 0 is x = 0 i ff i ff the columns of A are linearly independent. QED

Matrix invertibility examples  1 � 2 3 is not square so cannot be invertible. 4 5 6  1 � 2 is square and its columns are linearly independent so it is invertible. 3 4 2 1 1 2 3 5 is square but columns not linearly independent so it is not invertible. 2 1 3 4 3 1 4

Transpose of invertible matrix is invertible Theorem: The transpose of an invertible matrix is invertible. 2 a 1 3 2 3 2 3 . 4 v 1 A T = 4 a 1 5 = . v n a n A = · · · · · · 6 7 . 5 4 5 a n 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 A T are rows of A , so columns of A T are linearly independent. Since A T is square and columns are linearly independent, A T is invertible. QED

More matrix invertibility Earlier we proved: If A has an inverse A − 1 then AA − 1 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 . Then A 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 A u = 0 . Then B ( A u ) = B 0 = 0 . On the other hand, ( BA ) u = 1 u = u , so u = 0 . This shows A has an inverse A − 1 . Now must show B = A − 1 . We know AA − 1 = 1 . BA = 1 ( BA ) A − 1 = 1 A − 1 by multiplying on the right by A − 1 ( BA ) A − 1 = A − 1 B ( AA − 1 ) = A − 1 by associativity of matrix-matrix mult B 1 = A − 1 B = A − 1 QED

Representations of vector spaces Two important ways to represent a vector space: As Span { b 1 , . . . , b k } As the solution set of homogeneous linear system a 1 · x = 0 , . . . , a m · x = 0 Equivalently, 2 3 2 a 1 3 . 6 7 . Equivalently, Null 6 . 7 6 7 b 1 b k Col · · · 4 5 6 7 a m 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 a ffi ne spaces? I From representation as solution set of linear system to representation as a ffi ne hull I From representation as a ffi ne hull to representation as solution set of linear system

Conversions for a ffi ne spaces? From representation as solution set of linear system to representation as a ffi ne hull I input: linear system A x = b I output: vectors whose a ffi ne hull is the solution set of the linear system.  1 � 2 3 x  � 1 2 1 5 = y 4 − 1 2 2 2 z Let u be one solution to the linear system. u = [ − 0 . 5 , 0 . 75 , 0] 1  0 � 2 3 x  � 1 2 1 5 = Consider the corresponding homogeneous system A x = 0 . 2 y 4 − 1 2 2 0 z Its solution set, the null space of A , is a vector space V . 3 Let b 1 , . . . , b k be generators for V . b 1 = [2 , − 3 , 4] 4 Then the solution set of the original linear system is the a ffi ne hull of u , b 1 + u , b 2 + u , . . . , b k + u . [ − 0 . 5 , − . 75 , 0] and [ − 0 . 5 , − . 75 , 0] + [2 , − 3 , 4]

From representation as solution set to representation as a ffi ne hull  1 � 2 3 x  1 2 1 � 5 = is u = [ − 0 . 5 , 0 . 75 , 0] One solution to equation y 4 − 1 2 2 2 z  � 1 2 1 Solution set of equation is u + Span { b 1 } , is Span { b 1 } : Null space of − 1 2 2 i.e. the a ffi ne hull of u and u + b 1 b 1 u+b 1 u

Representations of vector spaces Two important ways to represent a vector space: As Span { b 1 , . . . , b k } As the solution set of homogeneous linear system a 1 · x = 0 , . . . , a m · x = 0 Equivalently, 2 3 2 a 1 3 . 6 7 . Equivalently, Null 6 7 . 6 7 Col b 1 · · · b k 4 5 6 7 a m 6 7 4 5

Representations of vector spaces Two important ways to represent a vector space: As Span { b 1 , . . . , b k } As the solution set of homogeneous linear system a 1 · x = 0 , . . . , a m · x = 0 Equivalently, 2 3 2 a 1 3 . 6 7 . Equivalently, Null 6 7 . 6 7 Col b 1 · · · b k 4 5 6 7 a m 6 7 4 5 How to transform between these two representations? Problem 1 (From left to right): I input: homogeneous linear system a 1 · x = 0 , . . . , a m · x = 0, I output: basis b 1 , . . . , b k for solution set Problem 2 (From right to left): I input: independent vectors b 1 , . . . , b k , I output: homogeneous linear system a 1 · x = 0 , . . . , a m · x = 0 whose solution set equals

Reformulating Problem 1 I input: homogeneous linear system a 1 · x = 0 , . . . , a m · x = 0, I output: basis b 1 , . . . , b k for solution set Let’s express this in the language of matrices: 2 a 1 3 . I input: matrix A = . 6 7 . 4 5 a m 2 3 4 b 1 I output: matrix B = b k · · · 5 such that Col B = Null A 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 1 · x = 0 , . . . , a m · x = 0, I output: basis b 1 , . . . , b k for solution set Let’s express this in the language of matrices: 2 a 1 3 . I input: matrix A = . 6 7 . 4 5 a m 2 3 with independent columns 4 b 1 I output: matrix B = b k · · · 5 such that Col B = Null A 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 1 · x = 0 , . . . , a m · x = 0, I output: basis b 1 , . . . , b k for solution set Let’s express this in the language of matrices: 2 a 1 3 . I input: matrix A = . 5 with independent rows 6 7 . 4 a m 2 3 with independent columns 4 b 1 I output: matrix B = b k · · · 5 such that Col B = Null A 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 the reformulation of Problem 1 I input: matrix A with independent rows I output: matrix B with independent columns such that Col B = Null A By Rank-Nullity Theorem, rank A + nullity A = n Because rows of A are linearly independent, rank A = m , so m + nullity A = n Requiring Col B = Null A is the same as requiring (i) Col B is a subspace of Null A (ii) dim Col B = nullity A

Reformulating the reformulation of Problem 1 I input: matrix A with independent rows I output: matrix B with independent columns such that Col B = Null A By Rank-Nullity Theorem, rank A + nullity A = n Because rows of A are linearly independent, rank A = m , so m + nullity A = n Requiring Col B = Null A is the same as requiring 2 3 0 · · · 0 . . . . (i) Col B is a subspace of Null A = ⇒ same as requiring AB = 6 7 . . 4 5 0 · · · 0 (ii) dim Col B = nullity A

Reformulating the reformulation of Problem 1 I input: matrix A with independent rows I output: matrix B with independent columns such that Col B = Null A By Rank-Nullity Theorem, rank A + nullity A = n Because rows of A are linearly independent, rank A = m , so m + nullity A = n Requiring Col B = Null A is the same as requiring 2 3 0 · · · 0 . . . . (i) Col B is a subspace of Null A = ⇒ same as requiring AB = 6 7 . . 4 5 0 · · · 0 (ii) dim Col B = nullity A = ⇒ same as requiring number of columns of B = nullity A same as requiring number of columns of B = n − m

Reformulating the reformulation of Problem 1 I input: matrix A with independent rows I output: matrix B with independent columns such that Col B = Null A By Rank-Nullity Theorem, rank A + nullity A = n Because rows of A are linearly independent, rank A = m , so m + nullity A = n Requiring Col B = Null A is the same as requiring 2 3 0 · · · 0 . . . . (i) Col B is a subspace of Null A = ⇒ same as requiring AB = 6 7 . . 4 5 0 · · · 0 (ii) dim Col B = nullity A = ⇒ same as requiring number of columns of B = nullity A same as requiring number of columns of B = n − m h 0 I input: m × n matrix A with independent rows i I output: matrix B with n − m independent columns such that AB =

Hypothesize a procedure for reformulation of Problem 1 Problem 1: h 0 I input: m × n matrix A with independent rows i I output: matrix B with n − m independent columns such that AB = Define procedure null space basis(M) with this spec: h 0 I input: r × n matrix M with independent rows i I output: matrix C with n − r independent columns such that MC =