# Basis and Dimension of matrices

If S is subspace of R6x6 consisting of all lower triangular matrices, what is the dimension of S?

Does anyone know the properties about dimension of lower triangular matrices?

Related Calculus and Beyond Homework Help News on Phys.org
Mark44
Mentor
The dimension of R6x6 is 36, right? One basis would consist of 36 matrices where each one has a single element of 1, and all other elements being 0. Each of the 36 matrices has the 1 element in a different place.

In your subspace, each matrix is guaranteed to have at least how many 0 elements, and where are they in each matrix?

Never mind. I think I found it :D

Editing: Yes, it is 21 :D

Mark44
Mentor
Yessireebob

Ok, I think this one is a little bit trickier
If S is the subspace of R5x5 consisting of all matrices with trace 0, then what is dimension of S?

I found out that trace is sum of the diagonal.
And I also found out that symmetric matrices have zero trace.

The problem is I don't know how to find the basis of 5x5 symmetric matrices?

Mark44
Mentor
Ok, I think this one is a little bit trickier
If S is the subspace of R5x5 consisting of all matrices with trace 0, then what is dimension of S?

I found out that trace is sum of the diagonal.
And I also found out that symmetric matrices have zero trace.
That's not true. A matrix A is symmetric if A = AT. The 2 x 2 identity matrix is symmetric, but its trace is 2. The 5 x 5 identity matrix is also symmetric, and its trace is 5.

So let's leave symmetric matrices out of this, since there is no mention of them in your problem statement. What single equation describes every matrix in your subspace S?
The problem is I don't know how to find the basis of 5x5 symmetric matrices?

you mean
ax1 + bx2 + cx3 + dx4 + ex5 = 0 ?

where a, b, c, d, e are elements on diagonal

Last edited:
Mark44
Mentor
Something like that. a1 + a2 + a3 + a4 + a5 = 0, where ai really means ai, i.

Here you have one equation with five unknowns. How many degrees of freedom do you have?

4? So the dimension is 4?

Mark44
Mentor
Right.

But when I submitted the answer, it said incorrect....

http://img192.imageshack.us/img192/6831/49103767.png [Broken]

Last edited by a moderator:
Mark44
Mentor
I guess I didn't take into account that we're dealing with 5 x 5 matrices. The space R5x5 has dimension 25. For your subspace of matrices whose trace is 0, you have 20 matrix elements off the main diagonal that are arbitrary, and 4 elements on the diagonal that are arbitrary. My logic might be flawed here, but I'm going to say that the dimension is 24.

It said 24 is correct.
So we were dealing with R5.
This is R5x5. I guess the number elements on the main diagonal should be more than 20, but I only see 5 elements: a11, b22, c33, d44, e55 of 5x5 matrices A, B, C, D, E.........

vela
Staff Emeritus
Homework Helper