Elementary Matrices

Definition An (n x n) matrix is an elementary matrix if it is obtained from the identity matrix by a single elementary row operation. Elementary row operations: Type I: Interchange two rows. Type II: Multiply a row by a non-zero constant. Type III: Add a multiple of one row to another.

Examples Type I Type II Type III

Notice result of multiplying EA Select 2x2 mtx of variables and multiply to show that could not possibly get I

Continued…

Summary of Effects To interchange two rows of matrix: Multiply the matrix times the elementary matrix of type I To multiply a row times a scalar: Multiply the matrix times the elementary matrix of type II To add a multiple of one row to another: Multiply the matrix times the elementary matrix of type III.

Theorem 1 If A is any (m x n) matrix and E the (m x m) elementary matrix obtained by performing an elementary row operation on the (m x m) identity. EA will result in the same matrix as you would get from performing the same row operation on A.

Example Find E such that EA is the result of adding 4 times row two to row 3:

Inverting the elem. row ops Type I: interchange rows p and q Type II: multiply row p by c≠0 Type III: add k times row p to row q Inverse ops Interchange rows p and q Multiply row p by 1/c Subtract k times row p from row q.

Theorem 2 Every E is invertible, and the inverse is an E of the same type: E-1 is the elementary mtx obtained from I by the inverse of the row operation that produced E from I. Proof: E is the result of performing row op k on I E’ is result of performing k’(inverse of k) on I Apply k to A and get EA Apply k’ to EA and get E’(EA) k’ reverses k, so applying k then k’ to A results in A So E’EA = A Choose A = I, then E’E = I. (Could show EE’ = I) �

Example Write inverse of following elementary matrices:

An extension Applying E1, E2,…,Ek to A gives EkEk-1…E2E1A Call U=EkEk-1…E2E1, then get UA Could also get U by applying the same sequence of row operations to I. So same sequence of row operations applied to [A I] results in [UA U] (becomes helpful later)

Theorem 3 Given that A is an (m x n) matrix, and A can be taken to B by sequence of elementary row operations: B = UA where U is an invertible (m x m) mtx U = EkEk-1…E2E1 where the E’s are the elem. Matrices corresponding to the elem. Row ops that take A to B (in reverse order). U can be constructed without finding Ei by taking [A I] to [UA U] (same ops that take A to UA take I to U)

Invertibility of U U is simply a product of elementary matrices, which are invertible by theorem 2. From 2.3 theorem 4 (part 4), the product of invertible matrices is also invertible. Therefore, U is invertible.

Example Put A into reduced row echelon form and then express it as UA (where U is the product of elementary matrices). Hint: since we know that the same sequence of elementary row operations that takes A to UA is the same as those which take I to U, we can simply: Augment A with I2, then row reduce. What we get on right is U.

Factoring U into E’s Take the same A matrix, and instead of finding U, find E’s such that U=Ek…E2E1 Hint: put A into rref noting the elem. Row ops you perform. Find E’s to match.

Theorem 4 The following are equivalent for (n x n) matrix A: A is invertible If YA = 0 where Y is (1 x n), then Y = 0 A has rank n. A can be carried to In by elementary row operations. A is a product of elementary matrices.

Proof of Theorem 4 To prove 5 statements equivalent, we simply show that 1implies 2 which implies 3, etc, and 5 implies 1 Proof: 1 implies 2: Given A is invertible, YA = 0 implies Y=YI= YAA-1 = 0A-1=0

Proof of Thm 4 (continued) 2 implies 3: we can take A to UA in reduced row echelon form (by thm 3) by a sequence of elem. Row ops. We just need to show that UA has n non-zero rows. If this is not true, then UA has a row of zeros at bottom. If we choose Y=[0 0 0 … 1], then Y(UA)=0 which implies YU = 0 (based on prop 2) Since U is invertible, Y=0 (from 2.3) (contradiction)

Proof of Thm 4 (continued) 3 implies 4: Given--A can be put into reduced row echelon form with n non-zero rows. Since the resulting matrix is (n x n) it will be In. 4 implies 5: I = UA where U=Ek…E2E1 and U is invertible (by thm 3). U-1UA = U-1I so A = U-1 =(Ek…E2E1)-1=E1-1E2-1…Ek-1 So A is a product of elementary matrices

Proof of Thm 4 (continued) 5 implies 1: A is a product of elementary matrices. Elementary matrices are all invertible. The product of invertible matrices is also invertible (by 2.3 theorem 4). Therefore, A is invertible.

Example Write A as a product of elementary matrices: Hint: Put in reduced row echelon form and determine E’s to match each elem. Row operation. Then Ek…E2E1A=I , so A = (Ek..E2E1)-1

Theorem 5 A and B are (n x n) matrices. If AB = I, then BA = I, so A and B are invertible and A = B-1, B=A-1. Proof: If YA = 0, then 0B = (YA)B=Y(AB)=YI=Y Since Y = 0, A is invertible. Since A is invertible, A-1AB = A-1I So B = A-1. 

Theorem 6 If A is (n x n), the following are equivalent A is invertible The homogeneous system AX = 0 has only the trivial solution X = 0. The system AX = B has a solution for every (n x 1) matrix B. Proof: 2.3 theorem 2 said that if A is (n x n) and invertible, then AX = B has a unique solution and X = A-1B. so 1 implies 2 and 3 above by this theorem

Proof of Theorem 6(cont) 2 implies 1: We try to show AT invertible first by using thm. 4 which says that if YAT = 0 implies Y=0, then AT is invertible: Assume YAT = 0 Then AYT = (YAT)T = 0T=0 By property 2, this means that YT = 0, so Y=0 Therefore, AT is invertible by theorem 4 Therefore, A is invertible by 2.3 theorem 4

Proof of Theorem 6(cont) 3 implies 1: prop 3 says that AX = B has a solution for every (n x 1) matrix B. So we can put together B= [B1 B2 … Bn], and X=[X1 X2 … Xn] such that the X’s are solutions matching the B’s. Then AX = B (put together) So if we define Bj = jth column of In then, [B1 B2 … Bn] = In AX = I, so A is invertible.