Suppose that M 0 = , M 1 = , M 2 = , M 3 = , Find the elementary matrices D 1, D
ID: 3103241 • Letter: S
Question
Suppose that M 0 = , M 1 = , M 2 = , M 3 = , Find the elementary matrices D 1, D 2, D 3 such that D 1 M 0 = M 1 for i = 1,2,3. For each positive integer n, let k(n) denote the smallest number with the following property: For any n times n lower triangular matrix A with nonzero digonal entries, there exist k(n) elementary matrices E 1, E 2, ,W k(n) such that E 1E 2 E k(n)A = I n. Prove that k(n) n(n + 1)/2. (Note: Informally, this question asks you to give an upper bound for how many elementary matrices are needed to ' invert ' an invertible lower triangular n times n matrix.)Explanation / Answer
a)
D1 =
1 0 0 0
0 0 0 1
0 0 1 0
0 1 0 0
The "scrambled" order of the 1s in this matrix reflect the same "scrambled" order of the rows in M1 as compared to M0
D2 =
1 0 0 0
0 1 0 0
0 0 3 0
0 0 0 1
In this case the value you 3 indicates the 3rd row of M2 is simply multiplied by 3 from M0
D3 =
1 0 0 0
2 1 0 0
0 0 1 0
0 0 0 1
This is a combination of ideas from parts 1 and 2
b)
I'm not exactly sure about the proof, but I know it is similar to the way that the sum of integers from 1 to n is (n)(n+1)/2
If you think about it, it's a similar idea. We are combining all these matrices, doing operations on them as in part (a), and coming up with the product of all these Elementary matrices being the inverse of A (so it multiplies to the Identity). Thus we will never need MORE than (n)(n+1)/2.