Matrix invertibility. Rank-Nullity Theorem: For any n-column matrix A, nullity A +ranka = n
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1 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 : F C! F R 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 =0anddimF C =dimf R 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
2 Matrix invertibility examples apple is not square so cannot be invertible. apple is square and its columns are linearly independent so it is invertible. 5 is square but columns not linearly independent so it is not invertible.
3 Transpose of invertible matrix is invertible Theorem: The transpose of an invertible matrix is invertible. a 1 v 1 v n A = 5 = 6. a n 5 A T = a 1 a n 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 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
4 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. 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. This shows A has an inverse A 1. Now must show B = A 1. We know AA 1 = 1. BA = 1 (BA)A 1 = 1A 1 by multiplying on the right by A 1 (BA)A 1 = A 1 B(AA 1 ) = A 1 B 1 = A 1 B = A 1 by associativity of matrix-matrix mult QED
5 Representations of vector spaces Two important ways to represent a vector space: As the solution set of homogeneous linear system a 1 x =0,...,a m x =0 Equivalently, Null 6 a 1. a m 5 As Span {b 1,...,b k } Equivalently, Col 6 b 1 b k 5
6 Conversions between the two representations {[x, y, z] : [, 1, 1] [x, y, z] =0, [0, 1, 1] [x, y, z] =0} Span {[1,, ]} Span {[, 1, 1], [0, 1, 1]} {[x, y, z] : [1,, ] [x, y, z] =0}
7 Conversions for a ne spaces? I From representation as solution set of linear system to representation as a ne hull I From representation as a ne hull to representation as solution set of linear system
8 Conversions for a ne spaces? From representation as solution set of linear system to representation as a ne hull I input: linear system Ax = b I output: vectors whose a ne hull is the solution set of the linear system. apple Let u be one solution to the linear system. u =[ 0.5, 0.5, 0] apple x 1 1 Consider the corresponding homogeneous system Ax = 0. y 1 z Its solution set, the null space of A, isavectorspacev. Let b 1,...,b k be generators for V. b 1 =[,, ] Then the solution set of the original linear system is the a ne hull of u, b 1 + u, b + u,...,b k + u. [ 0.5,.5, 0] and [ 0.5,.5, 0] + [,, ] x y z 5 = 5 = apple 1 apple 0 0
9 From representation as solution set to representation as a One solution to equation Null space of apple apple is Span {b 1 }: x y z 5 = apple 1 is u =[ 0.5, 0.5, 0] ne hull Solution set of equation is u +Span{b 1 }, i.e. the a ne hull of u and u + b 1 b 1 u+b 1 u
10 Representations of vector spaces Two important ways to represent a vector space: As the solution set of homogeneous linear system a 1 x =0,...,a m x =0 Equivalently, Null 6 a 1. a m 5 As Span {b 1,...,b k } Equivalently, Col 6 b 1 b k 5
11 Representations of vector spaces Two important ways to represent a vector space: As the solution set of homogeneous linear system a 1 x =0,...,a m x =0 Equivalently, Null 6 a 1. a m 5 As Span {b 1,...,b k } Equivalently, Col 6 b 1 b k 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 (From right to left): I input: independent vectors b 1,...,b k, I output: homogeneous linear system a x = 0,...,a x = 0 whose solution set equals
12 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: I input: matrix A = 6 I output: matrix B = a 1. a m 5 b k b 1 5 such that Col B =NullA 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.)
13 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: I input: matrix A = 6 I output: matrix B = a 1. a m 5 b k b 1 5 with independent columns such that Col B =NullA 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.)
14 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: I input: matrix A = 6 I output: matrix B = a 1. a m b k b with independent rows with independent columns such that Col B =NullA 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.)
15 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
16 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 =) same as requiring AB = (ii) dim Col B =nullitya
17 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 =) same as requiring AB = (ii) dim Col B =nullitya =) same as requiring number of columns of B =nullitya same as requiring number of columns of B = n m 5
18 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 =) same as requiring AB = (ii) dim Col B =nullitya =) same as requiring number of columns of B =nullitya same as requiring number of columns of B = n m I input: m n matrix A with independent rows h I output: matrix B with n m independent columns such that AB = 0 i 5
19 Hypothesize a procedure for reformulation of Problem 1 Problem 1: I input: m n matrix A with independent rows h I output: matrix B with n m independent columns such that AB = 0 i Define procedure null space basis(m) with this spec: I input: r n matrix M with independent rows h I output: matrix C with n r independent columns such that MC = 0 i
20 Reformulating Problem 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 Span {b 1,...,b k } Let s express this in the language of matrices: I input: n k matrix B with independent columns I output: matrix A with independent rows such that Null A = Col B As before, Rank-Nullity Theorem implies number of rows of A +nullitya = number of columns of A As before, requiring h Null A = Col B is the same as requiring (i) AB = 0 i (ii) number of rows of A = n k I input: n k matrix B with independent rows h I output: matrix A with n k independent rows such that AB = 0 i
21 Solving Problem with the procedure for Problem 1 Problem 1: I input: m n matrix A with independent rows h I output: matrix B with n m independent columns such that AB = 0 i Define procedure null space basis(m) I input: r n matrix M with independent rows h I output: matrix C with n r independent columns such that MC = 0 i Problem : I input: n k matrix B with independent rows h I output: matrix A with n k independent rows such that AB = 0 i To solve Problem, call null space basis(b T ). h Returns matrix A T with independent columns such that B T A T = 0 i Since B T is k nh matrix, A T has n k columns. Therefore AB = 0 i and A has n k independent rows. Therefore A is solution to Problem
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