Matrix inverses Recall... DefinitionMatrix inversesA square matrix A is invertible (or nonsingular) if matrix 9 BRecall...such that AB = I and BA = I. (We say B is an inverse of A.) Definition A square matrix A is invertible (or nonsingular) if matrix B Remark Not all square matrices are invertible. ∃ such that AB = I and BA = I. (We say B is an inverse of A.) Theorem.Remark IfNotA allis squareinvertible, matrices then are its invertible. inverse is unique. 1 RemarkTheorem.WhenIf A Ais invertible,is invertible, then we its denote inverse itsis unique. inverse as A− . 1 Remark When A is invertible, we denote its inverse as A− . Theorem. If A is an n n invertible matrix, then the system of × ~ 1~ linearTheorem. equationsIf A given is an n by A~xn invertible= b has matrix, the unique then solutionthe system~x of= linearA− b. × 1 equations given by Ax = b has the unique solution x = A− b. Proof. Assume A is an invertible matrix. Then we have by associativity of by def'n of Proof. Assume A is an invertiblematrix m matrix.ult. Theniden wetit havey 1 1 A(A− b)=(AA− )b = Ib = b. by def'n of inverse Theorem (Properties1 of matrix inverse). Thus, ~x = A− ~b is a solution to A~x = ~b. 1 1 1 Suppose(a) If A~y isis another invertible, solution then A to− theis itself linear invertible system. and It follows(A− )− that= AA~y. = ~b, 1 1 but multiplying both sides by A− gives ~y = A− ~b = ~x. (b) If A is invertible and c =0is a scalar, then cA is invertible and 1 1 1 (cA)− = c A− . Theorem (Properties of matrix inverse). (c) If A and B are both n n invertible matrices, then AB is invertible 1 1 1× 1 1 1 (a) IfandA is(AB invertible,)− = B− thenA− . A− is itself invertible and (A− )− = A. “socks and shoes rule” – similar to transpose of AB (b) IfgeneralizationA is invertible to product and c = of 0nismatrices a scalar, then cA is invertible and 1 6 1 1 T T 1 1 T (d)(cAIf )A− is= invertible,c A− . then A is invertible and (A )− =(A− ) . (c)To If proveA and (d),B weare need both ton shown thatinvertible the matrix matrices,B that then satisfiesAB is invertible T 1 T 1 ×1 1 T BAand=(IABand)−A=BB=−IAis−B.=(A− ) . LectureLecture 8 8 Math Math 40, 40, Spring Spring ’12,'12, Prof. Prof. Kindred Kindred Page Page 1 1 \socks and shoes rule" { similar to transpose of AB generalization to product of n matrices T T 1 1 T (d) If A is invertible, then A is invertible and (A )− = (A− ) . To prove (d), we need to show that the matrix B that satisfies T T 1 T BA = I and A B = I is B = (A− ) . 1 Proof of (d). Assume A is invertible. Then A− exists and we have 1 T T 1 T T (A− ) A = (AA− ) = I = I and T 1 T 1 T T A (A− ) = (A− A) = I = I: T T 1 1 T So A is invertible and (A )− = (A− ) . Recall... How do we compute the inverse of a matrix, if it exists? Inverse of a 2 2 matrix: Consider the special case where A is a × 2 2 matrix with A = [ a b ]. If ad bc = 0, then A is invertible and its × c d − 6 inverse is 1 1 d b A− = − : ad bc c a − − How do we find inverses of matrices that are larger than 2 2 matrices? × Theorem. If some EROs reduce a square matrix A to the identity matrix 1 I, then the same EROs transform I to A− . EROs -1 A I I A 1 If we can transform A into I, then we will obtain A− . If we cannot do so, then A is not invertible. Lecture 8 Math 40, Spring '12, Prof. Kindred Page 2 Can we capture the effect of an ERO through matrix multiplication? Definition An elementary matrix is any matrix obtained by doing an ERO on the identity matrix. Examples 0 1 0 0 1 0 4 R1 R2 1 0 0 0 R1 4R3 − on 4 $4 identity 2 3 on 3 −3 identity 0 1 0 × 0 0 1 0 × 2 3 6 7 0 0 1 60 0 0 17 6 7 4 5 Notice that 4 5 1 0 4 a a a a 4a a 4a a 4a − 11 12 13 11 − 31 12 − 32 13 − 33 0 1 0 a a a = a a a 2 3 2 21 22 233 2 21 22 23 3 0 0 1 a31 a32 a33 a31 a32 a33 4 5 4 5 4 5 Left mult. of A by row vector is a linear comb. of rows of A. 1 Remark An elementary matrix E is invertible and E− is elementary matrix corresponding to the \reverse" ERO of one associated with E. Example If E is 2nd elementary matrix above, then \reverse" ERO is 1 0 4 R + 4R and E 1 = 0 1 0 . 1 3 − 2 3 0 0 1 4 5 1 Remark When finding A− using Gauss-Jordan elimination of [ A I ], j if we keep track of EROs, and if E1;E2;:::;Ek are corresponding elem. matrices, then we have 1 1 1 EkEk 1 E1A = I = A = E1− Ek− 1Ek− : − ··· ) ··· − Lecture 8 Math 40, Spring '12, Prof. Kindred Page 3 Theorem (Fundamental Thm of Invertible Matrices). For an n n matrix, the following are equivalent: × (1) A is invertible. (2) A~x = ~b has a unique solution for any ~b Rn. 2 (3) A~x = ~0 has only the trivial solution ~x = 0. (4) The RREF of A is I. (5) A is product of elementary matrices. 1 5 2 Proof strategy 4 3 Proof. (1) (2): (2) (3): ) ) Proven in first theorem If A~x = ~b has unique sol'n for any of today's lecture ~b Rn, then in particular, A~x = ~0 2 has a unique sol'n. Since ~x = ~0 is a solution to A~x = ~0, it must be the unique one. (3) (4): (4) (5): ) ) If A~x = ~0 has unique sol'n ~x = 0, Ek E A = RREF of A = I ··· 1 then augmented matrix has no free and elem. matrices are invertible 1 1 1 variables and a leading one in every = A = E− E− E− : ) 1 ··· k 1 k column: − 1 0 1 0 2 . 3 (5) (1): .. ) 6 7 Since A = Ek E1 and Ei invertible 6 1 0 7 ··· 6 7 i, A is product of invertible matri- 4 5 8 so RREF of A is I. ces so it is itself invertible. Lecture 8 Math 40, Spring '12, Prof. Kindred Page 4 Theorem. Let A be a square matrix. If B is a square matrix such 1 that either AB = I or BA = I, then A is invertible and B = A− . Proof. Suppose A; B are n n matrices and that BA = I. Then consider × the homogeneous system A~x = ~0. We have B(A~x) = B~0 = (BA) ~x = ~0 = ~x = ~0: ) ) I Since A~x = ~0 has only the trivial solution ~x = ~0, by the Fundamental | {z } 1 Thm of Inverses, we have that A is invertible, i.e., A− exists. Thus, 1 1 1 1 1 (BA)A− = IA− = B (AA− ) = A− = B = A− : ) ) I We leave the case of AB = I as an exercise. | {z } n Definition The vectors ~e1;~e2; : : : ;~en R , where ~ei has a one in its 2 ith component and zeros elsewhere, are called standard unit vectors. Example The 4 4 identity matrix can be expressed as × 1 0 0 0 0 1 0 0 j j j j I = 2 3 = ~e ~e ~e ~e 4 0 0 1 0 2 1 2 3 43 6 7 60 0 0 17 j j j j 6 7 4 5 4 5 Theorem. If some EROs reduce a square matrix A to the identity 1 matrix I, then the same EROs transform I to A− . Why does this work? Want to solve AX = I, with X unknown n n matrix. × If ~x1; : : : ; ~xn are columns of A, then want to solve n linear systems A~x1 = ~e1; : : : ; A~xn = ~en. Can do so simultaneously using one \super-augmented matrix." Lecture 8 Math 40, Spring '12, Prof. Kindred Page 5.