# Dimension and base of a set

• Feb 19th 2009, 11:53 AM
arbolis
Dimension and base of a set
Hi MHF,
I'm unable to solve this problem. A little help would be welcome. (not a full answer if possible).
Find the dimension and a base of this set : $\{ A \in \mathbb{R}^{n\times n} \text {such that }A=A^T \}$.
My attempt : I'm tempted to answer $\dim S =n^2$ without thinking, but when I think I realize I don't know. $S$ is the set containing all the matrices nxn such that they are equal to their transposematrix. I'm not able to see further in order to find the dimension of $S$.
And once I know the dimension, I'll know how many vectors form a base of $S$.
Thank you very much in advance.
• Feb 19th 2009, 12:22 PM
clic-clac
Hi

So you're considering the subspace of symmetric matrices. Given a symmetric $n\times n$ matrix, how many "parameters" (=coefficients) do you think you must choose to completely define it?
• Feb 19th 2009, 12:43 PM
arbolis
Quote:

Originally Posted by clic-clac
Hi

So you're considering the subspace of symmetric matrices. Given a symmetric $n\times n$ matrix, how many "parameters" (=coefficients) do you think you must choose to completely define it?

2n?
• Feb 19th 2009, 12:51 PM
clic-clac
Mhhh in fact, you just have to choose the upper coefficients, diagonal included. The other ones will be equal to their symmetric coefficient through the diagonal.
So you have $1+2+...+n$ coefficients to choose, an that gives you the dimension.
• Feb 19th 2009, 01:20 PM
arbolis
Quote:

Originally Posted by clic-clac
Mhhh in fact, you just have to choose the upper coefficients, diagonal included. The other ones will be equal to their symmetric coefficient through the diagonal.
So you have $1+2+...+n$ coefficients to choose, an that gives you the dimension.

Thank you very much! Very well explained. So $\dim S =\frac{n(n+1)}{2}$.
I'll try now the other part of the question.
• Feb 19th 2009, 02:27 PM
arbolis
Checking my result. A base $\bold B$ is $\{ \begin{bmatrix} 1 & 0 & 0 & ... & 0 \\ 0 & 0 & 0 & ... & 0 \\ ...& ... & ... & ... & ... \\ 0 & 0 & 0 & ... & 0 \\ ...& ... & ... & ... & ... \end{bmatrix}, \begin{bmatrix} 0 & 1 & 0 & ... & 0 \\ 1 & 0 & 0 & ... & 0 \\ ...& ... & ... & ... & ... \\ 0 & 0 & 0 & ... & 0 \\ ...& ... & ... & ... & ... \end{bmatrix} ,$ $\begin{bmatrix} 0 & 0 & 0 & ... & 0 \\ 0 & 1 & 0 & ... & 0 \\ ...& ... & ... & ... & ... \\ 0 & 0 & 0 & ... & 0 \\ ...& ... & ... & ... & ... \end{bmatrix}, ...\}$. All matrices are nxn and there are $\frac{n(n+1)}{2}$ matrices forming the base. Am I right?
• Feb 19th 2009, 02:36 PM
ThePerfectHacker
For $1\leq i\leq j\leq n$ let $\bold{e}_{ij}$ be the matrix whose $ij$-th and $ji$-th entry is $1$ while everything else is zero. Notice if $i=j$ then $\bold{e}_{ij}$ is a matrix with only $1$ on the $ii$ (or $jj$) location on the diagnol.

For example, $n=3$ then $\bold{e}_{12}$ is:
$\begin{bmatrix}0&1&0\\1&0&0\\0&0&0\end{bmatrix}$

The basis is, $B = \{ \bold{e}_{ij} | 1\leq i\leq j\leq n\}$ and of course $|B| = \tfrac{1}{2}n(n+1)$.
• Feb 19th 2009, 03:24 PM
math2009
$A=\begin{bmatrix}a_{11} & \cdots & a_{ji} \\ \vdots & \ddots & \ \\ a_{ij} & \ & a_{nn} \end{bmatrix}$

$A=A^T$ , it means $a_{ij}=a_{ji}$. If $i\neq j$ , except diagonal,there are $n^2-n$ elements, $\because$ they are symmetric, $\therefore$ there are $\frac{n^2-n}{2}+n(diagonal)=\frac{n^2+n}{2}$ different elements. $\dim (A)=\frac{n^2+n}{2}$

$\mathfrak{B}=\{\begin{bmatrix}\ & \cdots & a_{ji} \\ \vdots & \ddots & \ \\ a_{ij} & \cdots & \ \end{bmatrix} | 1\leq i , j\leq n,a_{ij}=a_{ji}=1\}$
• Feb 19th 2009, 03:45 PM
arbolis
Ok, this is what I meant, but didn't know how to explain as TPH did and didn't know how to write it as math2009 did.
Thank you very much for the help!
• Feb 19th 2009, 03:55 PM
math2009
At begin, I didn't know how to write perfect math formula.
I check http://www.mathhelpforum.com/math-help/latex-help/ (LaTex Tutorial)
And apply WinEdt , then I know LaTex syntax.
You could click formula ,it pop up LaTex command.