Matrix Operation

1. Feb 4, 2009

Nusc

1. The problem statement, all variables and given/known data

If S is a lower triangular matrix (n by n), S^n is the zero matrix.

2. Relevant equations

3. The attempt at a solution

S =

0 0 0 0 ... 0
1 0 0 0 ... 0
0 1 0 0 ... 0
0 0 1 0 ... 0
...
0 0 0 0 .1.0

S^2 =

0 0 0 0 ... 0
0 0 0 0 ... 0
1 0 0 0 ... 0
0 1 0 0 ... 0
...
0 0 0 0 1.. 0

How do I show this using induction?

2. Feb 4, 2009

descendency

Are you sure it's true?

3. Feb 4, 2009

Nusc

No. So given S as I defined above then computing S^2, and S^3 so forth, the entries lower each row for each n. how do I generalize this?

4. Feb 4, 2009

descendency

Use the definition of matrix multiplication to show that the value of Sk-1k,1 = 1 but Sk-1< k,* = 0 and Sk-1k, >1and<n = 0 => Sk*,* = 0.

Where * means any row and any column.

edit: If you are trying to prove that statement in general, you might want to consider the identity matrix which is both n x n and lower triangular.

5. Feb 5, 2009

Staff: Mentor

First, a lower triangular matrix is one where the entries above the diagonal are zero. The matrix you show has ones down the lower subdiagonal and zeroes everywhere else.
Second, the statement isn't true.

Here's a counterexample for n = 3.

S =
[1 0 0]
[1 1 0]
[1 1 1]

S^2=
[1 0 0]
[2 1 0]
[3 2 1]

S^3=
[1 0 0]
[3 1 0]
[6 3 1]

The upshot is that S is a lower triangular 3 x 3 matrix, but S^3 != 0.

6. Feb 5, 2009

gabbagabbahey

Are you sure the question isn't in regards to a 'strictly lower triangular matrix'?

7. Feb 5, 2009

Nusc

The matrix is this:

S =

0 0 0 0 ... 0
1 0 0 0 ... 0
0 1 0 0 ... 0
0 0 1 0 ... 0
...
0 0 0 0 .1.0

it's not a lower triangular matrix.

8. Feb 5, 2009

Staff: Mentor

It would be helpful if we knew exactly how the problem is stated. In your first post, here is what you had:
The somewhat sparse n x n matrix with 1s on the subdiagonal appeared in your attempt at a solution.

Is the problem statement to show that for that matrix S, S^n = 0?

Inquiring minds would like to know...

9. Feb 5, 2009

Nusc

Okay.
Given
=

0 0 0 0 ... 0
1 0 0 0 ... 0
0 1 0 0 ... 0
0 0 1 0 ... 0
...
0 0 0 0 .1.0

Compute (S^n)(S^n*) and (S^*n)(S^n)

where * is the complex conjugate transpose.

10. Feb 6, 2009

HallsofIvy

Staff Emeritus
Actually, that is a "lower triangular matrix" (it has only 0s above the main diagonal), just not a general lower triangular matrix as was intially implied.

11. Feb 7, 2009

Nusc

So in Compute (S^n)(S^n*) and (S^*n)(S^n)
they're both zero correct?