Show that the subspace of the set of upper triangular n x n matrices of M(nn) has dimension (n^2 + n)/2 for n = 3

How do I set this up? Do I use simplified span or independent test method, and how?

Thanks.

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- April 15th 2012, 11:21 AMCantubaDimension of upper-triangular matrices
Show that the subspace of the set of upper triangular n x n matrices of M(nn) has dimension (n^2 + n)/2 for n = 3

How do I set this up? Do I use simplified span or independent test method, and how?

Thanks. - April 15th 2012, 12:41 PMuasacRe: Dimension of upper-triangular matrices
Hi,

The answer to your question is encoded in the definition of "dimension of the set of upper triangular matrices". You should count the number of independent variables that there exist in such matrices. - April 15th 2012, 02:24 PMDevenoRe: Dimension of upper-triangular matrices
prove that the matrices where is the matrix with 1 in the i,j-th position, and 0's elsewhere, form a basis for i ≤ j.

these matrices are clearly linearly independent, since they are a subset of a basis for Mat(n,F). showing they span the set of upper-triangular matrices is likewise fairly trivial.

so....how many of these matrices are there?

well there's 1 for every diagonal entry position, so n of these,

1 for every super-diagonal entry position, so n-1 more,

1 for every super-super diagonal position, so n-2 more,

....and so on until we get the the "last" super-diagonal, the upper-right corner, with one position.

add these all up: we have 1 + 2 +...+ n-1 + n = n(n+1)/2.

to see what i mean, in the space of 3x3 matrices, i am talking about the following set:

which has 1+2+3 = 6 elements.