Basis and Dimension of matrices

In summary, the dimension of a subspace S of R5x5 consisting of all matrices with trace 0 is 24. This is because the set of 5x5 matrices has a dimension of 25, but the condition of having a trace of 0 removes one degree of freedom, leaving a total of 24 degrees of freedom for the matrix in this subspace.
  • #1
lypaza
18
0
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?
 
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  • #2
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
Never mind. I think I found it :D

Editing: Yes, it is 21 :D
 
  • #5
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
lypaza said:
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?
lypaza said:
The problem is I don't know how to find the basis of 5x5 symmetric matrices?
 
  • #7
you mean
ax1 + bx2 + cx3 + dx4 + ex5 = 0 ?

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

http://img192.imageshack.us/img192/6831/49103767.png [Broken]
 
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  • #12
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
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
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
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.
 

1. What is the basis of a matrix?

The basis of a matrix is a set of linearly independent vectors that span the vector space of the matrix. It is the smallest set of vectors that can be used to express all other vectors in the vector space.

2. How is the dimension of a matrix determined?

The dimension of a matrix is determined by the number of linearly independent columns or rows it contains. It is represented by the number of entries in a basis for the matrix.

3. Can a matrix have more than one basis?

Yes, a matrix can have multiple bases as long as they both satisfy the conditions of being linearly independent and spanning the vector space of the matrix.

4. How does the basis of a matrix affect its rank?

The basis of a matrix does not directly affect its rank. However, the dimension of the vector space spanned by the basis is equal to the rank of the matrix. So, a larger basis can lead to a higher rank matrix.

5. Is the basis of a matrix unique?

No, the basis of a matrix is not unique. There can be multiple sets of linearly independent vectors that span the same vector space. However, the number of vectors in the basis will always be the same and will determine the dimension of the matrix.

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