A problem with basis and dimension

In summary, the conversation discusses the problem of finding the dimension of a subset P of Rnxn, where P is the set of antisymmetric matrices. The conversation also explores the number of independent components needed to fully describe an arbitrary antisymmetric nxn matrix, and arrives at the conclusion that the dimension of P is (n^2 - n)/2.
  • #1
chefobg57
8
0
Hi!

I am working on the following problem:

If a matrix is antisymmetric (thus A^T = -A), show that

P = {A [tex]\in[/tex] R | A is antisymmetric} is a subset of Rnxn. Also, find the dimension of P.

So far, I've proven that P is a subset of R and I am guessing that, in order for me to find the dimension, I need to find the basis of P first. Here is where I am kind of stuck.

I understand what the properties of the base would be (1. the vectors inside the set will be linearly independent and 2. the basis will be a spanning set for P), but how exactly should I start working so that I can find the basis itself...

A hint would be highly appreciated!

Thanks a bunch guys!
 
Physics news on Phys.org
  • #2
How many numbers do you need to fully describe an arbitrary antisymmetric nxn matrix?

In other words, how many independent components does such matrix have?
 
  • #3
Hm..
Well, I am debating between n and 2n numbers.
The following 2x2 matrix is antisymmetric:
0 a
-a 0

So, it depends on one number, a.

A description of a 3x3 matrix would depend on 3 numbers, a, b, and c and so forth.

Does this mean that this is the answer to the problem? n elements?

Thank you in advance!
 
  • #4
Try to write down an asymmetric 4x4 matrix.
 
  • #5
Thanks for pointing that out.

For 2x2 matrices,1 element is necessary.
For 3x3 matrices, 3 elements, for 4x4 - 6 elements, for 5x5 - 9 elements, for 6x6 it's 15 elements..

I am trying to come up with the proper function that will generate such output, given the input (n), but ... I can't really think of anything..
 
  • #6
chefobg57 said:
Thanks for pointing that out.

For 2x2 matrices,1 element is necessary.
For 3x3 matrices, 3 elements, for 4x4 - 6 elements, for 5x5 - 9 elements, for 6x6 it's 15 elements..

I am trying to come up with the proper function that will generate such output, given the input (n), but ... I can't really think of anything..
Consider that, say, a 6x6 matrix has 6 elements on the diagonal, 15 elements above the diagonal, and 15 elements below the diagonal. 6+15+15 = 36.
 
  • #7
Oh great!
So can we say then that the dimension for P would be:
(n^2 - n)/2?

Thank you so much!
 

FAQ: A problem with basis and dimension

What is a basis in linear algebra?

A basis in linear algebra is a set of linearly independent vectors that span the entire vector space. This means that any vector in the space can be expressed as a unique linear combination of the basis vectors.

What is the dimension of a vector space?

The dimension of a vector space is the number of vectors in a basis for that space. It represents the number of independent directions or degrees of freedom in the space.

Can a vector space have more than one basis?

Yes, a vector space can have multiple bases as long as they all have the same number of vectors, or the same dimension. This is because any basis for a vector space must have the same number of vectors, and any set of linearly independent vectors that span the space can serve as a basis.

What is the relationship between basis and dimension?

The dimension of a vector space is equal to the number of vectors in a basis for that space. This means that the dimension determines the number of basis vectors needed to span the entire space.

How do you determine the dimension of a vector space?

The dimension of a vector space can be determined by finding the number of linearly independent vectors in any basis for that space. This can be done by using row reduction to find the rank of the matrix representation of the vectors in the basis.

Back
Top