Linear Algebra Subspace Basis Problem

In summary, the conversation discusses finding a basis for the subspace of traceless (nxn)-matrices, denoted as sl(n), and determining its dimension. The dimension of sl(n) is n^2-n, as it is a subspace of (bold)K^(nxn) and is defined by the condition of having a zero sum of diagonals. The basis for this subspace can be constructed by combining the basis for matrices with all zeros down the diagonal and the basis for matrices that are zero off the diagonal but have nonzero elements along the diagonal. Together, these bases have a dimension of n^2, completing the basis for sl(n).
  • #1
braindead101
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1. The set of all traceless (nxn)-matrices is a subspace sl(n) of (bold)K^(nxn). Find a basis for sl(n). What is the dimension of sl(n)?

Not sure how to go about finding the basis. I know a basis is a list of vectors that is linearly independent and spans.
and for the dimension of sl(n), is it just n^2, as it's an nxn matrix.

any help would be greatly appreciated.
 
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  • #2
A traceless matrix has to satisfy exactly one extra condition relative to a general matrix. What does this tell you about it's dimension? Start by describing n^2-n matrices that have zeros down the diagonal. Then describe the matrices that have only diagonal elements but are still traceless.
 
  • #3
why is it n^2-n matrices?
I only know that traceless is where the sum of the diagonals is zero.
and I also have the following information from another problem:

the set sl(n) of all traceless (nxn)-matrices form a subspace of (bold)K^(nxn):
the zero matrix belongs to sl(n) so sl(n) is not empty.
tr(cA+B) = tr(cA) + tr(B)
tr(cA+B) = c tr(A) + tr(B)
tr(cA+B) = c (0) + 0 = 0
and cA+B (epsilon) sl(n) which means sl(n) is a subspace.
 
  • #4
A basis for the subspace of matrices that have all zeros down the diagonal has dimension n^2-n. Why? Construct an explicit basis. Now think about the matrices that are zero off the diagonal and only have nonzero elements along the diagonal but are traceless. Just sort of 'add the two'.
 
  • #5
okay, i understand now why the dimension is n^2-n
but do we need the dimension to find the basis. how do i start with the basis, we were given examples in class of our "favorite bases", and i remember the prof saying that we need them to find other bases.
 
  • #6
Only the dimension of the off diagonal elements is n^2-n. You have to add some elements to complete the basis for sl(n). They are purely diagonal. How many are there?
 

1. What is a subspace in linear algebra?

A subspace in linear algebra is a subset of a vector space that is closed under vector addition and scalar multiplication. This means that any linear combination of vectors in the subspace will also be in the subspace.

2. What is a basis in linear algebra?

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

3. How do you determine if a set of vectors form a basis for a subspace?

In order for a set of vectors to form a basis for a subspace, they must satisfy two conditions: they must span the subspace, meaning that any vector in the subspace can be written as a linear combination of the basis vectors, and they must be linearly independent, meaning that no vector in the set can be expressed as a linear combination of the other vectors.

4. What is the Linear Algebra Subspace Basis Problem?

The Linear Algebra Subspace Basis Problem is a mathematical problem that involves finding a basis for a given subspace. This problem is important in linear algebra because it allows us to represent vectors in a more concise and efficient way, and it also helps us better understand the structure and properties of vector spaces.

5. How do you solve the Linear Algebra Subspace Basis Problem?

The first step in solving the Linear Algebra Subspace Basis Problem is to determine if the given set of vectors span the subspace. If they do not, then additional vectors must be added to the set until it spans the subspace. Next, the linear independence of the set must be checked. If the set is not linearly independent, then redundant vectors must be removed. Once the set of vectors satisfies both conditions, it can be considered a basis for the subspace.

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