# Thread: Dimension and base of a set

1. ## 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.

2. 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?

3. 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?

4. 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.

5. 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.

6. 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?

7. 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 $|B| = \tfrac{1}{2}n(n+1)$.

8. $A=\begin{bmatrix}a_{11} & \cdots & a_{ji} \\ \vdots & \ddots & \ \\ a_{ij} & \ & a_{nn} \end{bmatrix}$

$A=A^T$ , 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. $\dim (A)=\frac{n^2+n}{2}$

$\mathfrak{B}=\{\begin{bmatrix}\ & \cdots & a_{ji} \\ \vdots & \ddots & \ \\ a_{ij} & \cdots & \ \end{bmatrix} | 1\leq i , j\leq n,a_{ij}=a_{ji}=1\}$

9. 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!

10. 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.