# Dimension of upper-triangular matrices

• Apr 15th 2012, 11:21 AM
Cantuba
Dimension 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.
• Apr 15th 2012, 12:41 PM
uasac
Re: Dimension of upper-triangular matrices
Hi,

The answer to your question is encoded in the definition of "dimension of the set of upper triangular $\displaystyle n\times n$ matrices". You should count the number of independent variables that there exist in such matrices.
• Apr 15th 2012, 02:24 PM
Deveno
Re: Dimension of upper-triangular matrices
prove that the matrices $\displaystyle \{E_{ij}\}$ where $\displaystyle E_{ij}$ 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:

$\displaystyle \left\{ \begin{bmatrix}1&0&0\\0&0&0\\0&0&0 \end{bmatrix},\ \begin{bmatrix}0&1&0\\0&0&0\\0&0&0 \end{bmatrix},\ \begin{bmatrix}0&0&1\\0&0&0\\0&0&0 \end{bmatrix},\ \begin{bmatrix}0&0&0\\0&1&0\\0&0&0 \end{bmatrix},\ \begin{bmatrix}0&0&0\\0&0&1\\0&0&0 \end{bmatrix},\ \begin{bmatrix}0&0&0\\0&0&0\\0&0&1 \end{bmatrix}\right\}$

which has 1+2+3 = 6 elements.