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Basis and Dimension of matrices

  • Thread starter lypaza
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  • #1
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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?
 

Answers and Replies

  • #2
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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?
 
  • #3
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Never mind. I think I found it :D

Editing: Yes, it is 21 :D
 
  • #4
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Yessireebob
 
  • #5
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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?
 
  • #6
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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?
 
  • #7
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you mean
ax1 + bx2 + cx3 + dx4 + ex5 = 0 ?

where a, b, c, d, e are elements on diagonal
 
Last edited:
  • #8
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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?
 
  • #9
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4? So the dimension is 4?
 
  • #10
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Right.
 
  • #11
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But when I submitted the answer, it said incorrect....

http://img192.imageshack.us/img192/6831/49103767.png [Broken]
 
Last edited by a moderator:
  • #12
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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.
 
  • #13
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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.........
 
  • #14
vela
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With an nxn real matrix, you start with n2 degrees of freedom since you have n2 entries in the matrix. Each independent equation you have for the elements removes one degree of freedom.

While the condition for the trace only involves elements on the diagonal, but those aren't the only ones that matter. You still have the ability to arbitrarily set the 20 off-diagonal elements, so you have a total of 20+4 degrees of freedom for the matrix as a whole.
 
  • #15
HallsofIvy
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Different way of seeing the same thing Vela said: the dimension of the set of 5 by 5 matrices is 25. Saying that the trace is 0 puts one constraint on that: the dimension of the set of 5 by b matrices with trace 0 is 25- 1= 24.
 

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