1. Prove Cases 2 and 3 in the Theorem about the product
EA
Case 2.
Suppose that E is obtained by multiplying the i-th row
of Im by a non-zero number, x. Then each row ra except ri has a
leading 1 on the (a,a) place and zeros everywhere else. The row ri has
the non-zero number x on the (i,i)-place and zeros everywhere else. If
a is not equal to i then the element EA(a,b), the product of ra and
cb, will be equal to the a-th entry of cb, which is A(a,b). Therefore
EA(a,b)=A(a,b) whenever a is not equal to i. Thus all rows of EA
except for the i-th row coincide with the corresponding rows of A. The
i-th row is the only one to change. The entries of this row are
products of ri and cb (b=1,...,n). The i-th row of E has x on the i-th
place and zeros everywhere else So the product of ri and cb is x times
the i-th element of cb. Thus EA(i,b)=xA(i,b) for every b=1,2,...,n.
Therefore the i-th row of EA is obtained by multiplying the i-th row of
A by x. This is the same operation we used to obtain E from i. The
proof is complete.
Case 3.
Suppose that E is obtained by
swapping the i-th and j-th rows of Im. Then each row ra except ri and
rj has a 1 on the (a,a) place and zeros everywhere else. ri has the 1
on the (i,j) place and zeros everywhere else. rj has the 1 on the
(j,i) place and zeros everywhere else. If a is not equal to i or j,
then the element EA(a,b), the product of ra and cb will be equal to
the a-th entry of cb, that is A(a,b). Therefore EA(a,b) = A(a,b)
whenever a is not equal to i or j and all rows but the i-th and j-th
rows coincide with the corresponding rows of A. The i-th and j-th rows
are the only ones to change. The entries of ri are the products of ri
and cb (b = 1...n). ri has a 1 on the j-th element and zeros
everywhere else. So the product of ri and cb is simply the j-th
element of cb. EA(i,b) = A(j,b) for every b = 1...n. Therefore, the
i-th row of EA is the j-th row of A. We can prove in the same way that
the j-th row of EA is the i-th row of A. The i-th and j-th rows of A are
swapped. This is the same row operation that performed to obtain E
from i. The proof is complete.
2. Express the matrix
| A := |
[ 1 |
0 |
1 ] |
| [ 0 |
1 |
1 ] |
| [ 1 |
1 |
0 ] |
as a product of elemetary matrices.
To do this, we first work backwards. We reduce A to the identity
matrix. This takes five row operations. To get matrix A from the
identity matrix, we perform the inverse operations in opposite order.
These five row operations are
- add row 3 to row 1 2) add -1*row 3 to row 2
- multiply row 3 by 2
- add row 2 to row 3
- add row 1 to row 2
So we have
five elementary matrices (E1, E2, E3, E4, and E5) and B, the identity
matrix I3.
> B:=matrix(3,3,[[1,0,0],[0,1,0],[0,0,1]]);
| B := |
[ 1 |
0 |
0 ] |
| [ 0 |
1 |
0 ] |
| [ 0 |
0 |
1 ] |
> E1:=addrow(B,3,1);
| E1 := |
[ 1 |
0 |
1 ] |
| [ 0 |
1 |
0 ] |
| [ 0 |
0 |
1 ] |
> E2:=addrow(B,3,2,-1);
| E2 := |
[ 1 |
0 |
0 ] |
| [ 0 |
1 |
-1 ] |
| [ 0 |
0 |
1 ] |
> E3:=mulrow(B,3,2);
| E3 := |
[ 1 |
0 |
0 ] |
| [ 0 |
1 |
0 ] |
| [ 0 |
0 |
2 ] |
> E4:=addrow(B,2,3);
| E4 := |
[ 1 |
0 |
0 ] |
| [ 0 |
1 |
0 ] |
| [ 0 |
1 |
1 ] |
> E5:=addrow(B,1,2);
| E5 := |
[ 1 |
0 |
0 ] |
| [ 1 |
1 |
0 ] |
| [ 0 |
0 |
1 ] |
The product of these elementary matrices (E5, E4, E3, E2, E1) is A.
> evalm(E5&*E4&*E3&*E2&*E1);
| [ 1 |
0 |
1 ] |
| [ 1 |
1 |
0 ] |
| [ 0 |
1 |
1 ] |
We have checked this using Maple, and the resulting matrix is A.