1. Elementary Matrices

2. 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.

3. Examples
Type I
Type II
Type III

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

5. Continued…

6. 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.

7. 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.

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

9. 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.

10. 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) �

11. Example
Write inverse of following elementary matrices:

12. 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)

13. 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)

14. 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.

15. 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.

16. 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.

17. 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.

18. 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

19. 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)

20. 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

21. 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.

22. 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

23. 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. 

24. 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

25. 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

26. 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.