35

Let $\textbf A$ denote the space of symmetric $(n\times n)$ matrices over the field $\mathbb K$, and $\textbf B$ the space of skew-symmetric $(n\times n)$ matrices over the field $\mathbb K$. Then $\dim (\textbf A)=n(n+1)/2$ and $\dim (\textbf B)=n(n-1)/2$.

Short question: is there any short explanation (maybe with combinatorics) why this statement is true?

EDIT: $\dim$ refers to linear spaces.

Rodrigo de Azevedo
  • 18,977
  • 5
  • 36
  • 95
Christian Ivicevic
  • 2,993
  • 3
  • 26
  • 40

5 Answers5

35

All square matrices of a given size $n$ constitute a linear space of dimension $n^2$, because to every matrix element corresponds a member of the canonical base, i.e. the set of matrices having a single $1$ and all other elements $0$.

The skew-symmetric matrices have arbitrary elements on one side with respect to the diagonal, and those elements determine the other triangle of the matrix. So they are in number of $(n^2-n)/2=n(n-1)/2$, ($-n$ to remove the diagonal).

For the symmetric matrices the reasoning is the same, but we have to add back the elements on the diagonal: $(n^2-n)/2+n=(n^2+n)/2=n(n+1)/2$.

enzotib
  • 9,612
  • 2
  • 21
  • 35
19

Here is my two cents:


\begin{eqnarray} M_{n \times n}(\mathbb{R}) & \text{has form} & \begin{pmatrix} *&*&*&*&\cdots \\ *&*&*&*& \\ *&*&*&*& \\ *&*&*&*& \\ \vdots&&&&\ddots \end{pmatrix} \hspace{.5cm} \text{with $n^2$ elements}\\ \\ \\ Skew_{n \times n}(\mathbb{R}) & \text{has form} & \begin{pmatrix} 0&*'&*'&*'&\cdots \\ *&0&*'&*'& \\ *&*&0&*'& \\ *&*&*&0& \\ \vdots&&&&\ddots \end{pmatrix} \end{eqnarray} For this bottom formation the (*)s and (*')s are just operationally inverted forms of the same number, so the array here only takes $\frac{(n^2 - n)}{2}$ elements as an argument to describe it. This appears to be an array geometry question really... I suppose if what I'm saying is true, then I conclude that because $\dim(Skew_{n \times n}(\mathbb{R}) + Sym_{n \times n}(\mathbb{R})) = \dim(M_{n \times n}(\mathbb{R}))$ and $\dim(Skew_{n \times n}(\mathbb{R}))=\frac{n^2-n}{2}$ then we have that \begin{eqnarray} \frac{n^2-n}{2}+\dim(Sym_{n \times n}(\mathbb{R})))=n^2 \end{eqnarray} or \begin{eqnarray} \dim(Sym_{n \times n}(\mathbb{R})))=\frac{n^2+n}{2}. \end{eqnarray}

Trancot
  • 3,953
  • 1
  • 25
  • 57
8

This is a bit late, but I figured I would chime in since the other answers don't really give the combinatorial intuition asked for in the original question.

In order to specify a skew-symmetric matrix, we need to choose a number $A_{ij}$ for each set $\{i,j\}$ of distinct indices. How many such sets are there? Combinatorics tells us that there are $\left( \begin{array}{@{}c@{}} n \\ 2 \end{array} \right) = \frac{n(n-1)}{2}$ such sets.

Similarly, in order to specify a symmetric matrix, we need to choose a number $A_{ij}$ for each set $\{i,j\}$ of (not necessarily distinct) indices. We can encode such sets by adding a special symbol $n+1$ that indicates a repeated index. Thus, we need to choose a number $A_{ij}$ for each set $\{i,j\}$ of distinct symbols, where now a symbol is either an index ($1 , \ldots , n$), or our special symbol $n+1$ that is used for a repeated index. Combinatorics tells us that there are $\left( \begin{array}{@{}c@{}} n+1 \\ 2 \end{array} \right) = \frac{n(n+1)}{2}$ such sets.

Stirling
  • 499
  • 3
  • 10
3

The dimension of symmetric matrices is $\frac{n(n+1)}2$ because they have one basis as the matrices $\{M_{ij}\}_{n \ge i \ge j \ge 1}$, having $1$ at the $(i,j)$ and $(j,i)$ positions and $0$ elsewhere. For skew symmetric matrices, the corresponding basis is $\{M_{ij}\}_{n \ge i > j \ge 1}$ with $1$ at the $(i,j)$ position, $-1$ at the $(j,i)$ position, and $0$ elsewhere.

Note that the diagonal elements of skew symmetric matrices are $0$, hence their dimension is $n$ less than the dimension of normal symmetric matrices.

Rijul Saini
  • 2,074
  • 16
  • 22
3

There have been many very good answers, so this is going to be another perspective of counting. We tackle the case of skew-symmetric first, which yields the condition that for any matrix $A$ with real entries, $a_{ij} = -a_{ij}$, so this means one side of the matrix is completely determined by the other as shown.

$$\begin{matrix} \begin{pmatrix} 0 &*' \\ *& 0 \\ \end{pmatrix} & * \end{matrix}$$

$$\begin{matrix} \begin{pmatrix} 0 &*' & *' \\ *& 0 & *' \\ * & * & 0 \\ \end{pmatrix} & \begin{matrix} *& \\ * & * \end{matrix} \end{matrix}$$

$$\begin{matrix} \begin{pmatrix} 0 &*' & *' &*'\\ *& 0 & *' &*'\\ * & * & 0 &*'\\ * & * &* & 0 \end{pmatrix} & \begin{matrix} * & & \\ *&* & \\ * & * &* \end{matrix} \end{matrix}$$

$$\begin{matrix} \begin{pmatrix} 0&*'&*'&*'&\cdots \\ *&0&*'&*'& \\ *&*&0&*'& \\ *&*&*&0& \\ \vdots&&&&\ddots \end{pmatrix} & \begin{matrix} *& \\ *&*& \\ *&*&*& \\ \vdots&&&&\ddots \end{matrix} \end{matrix}$$

Notice that the triangle keeps getting larger and larger and at each stage and is incremented by $+1$. Since we have $n-1$ entries to consider we add up the number of entries, that is $$1+ 2 + \dots + (n-1).$$

So by Gauss, we have $$\frac{(n-1)(n-1+1)}{2} = \frac{(n-1)n}{2}$$ degree of freedom and that is the dimension we seek.

For the symmetric case we need to add back the $n$ objects in diagonal zeroed out, so $$\frac{(n-1)n}{2} + n = \frac{n^2 + n}{2}$$

IAmNoOne
  • 3,124
  • 4
  • 16
  • 22