# Dimensions of symmetric and skew-symmetric matrices

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.

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Do you mean symmetric (not normal) in the title? And do you mean the $\dim$ of linear spaces of such matrices, not the $\dim$ of the matrices, right? – enzotib Aug 23 '12 at 9:59
I did edit it - thanks for the reminder! – Christian Ivicevic Aug 23 '12 at 10:02
You did not edit it correctly; $\mathbf A$ still refers to just one matrix, not a subspace. I will edit it for you. – Marc van Leeuwen Aug 23 '12 at 12:38
And you should say that $\mathbb K$ is not of characteristic $2$, or otherwise symmetric and anti-symmetric matrices are the same thing and your equations cannot both be true. – Marc van Leeuwen Aug 23 '12 at 12:40

## 3 Answers

All square matrices of a given size $n$ constitute a linear space of dimension $n^2$, because every matrix element is a component in the canonical base, ie. 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 of 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 the elements on the diagonal: $(n^2-n)/2+n=(n^2+n)/2=n(n+1)/2$.

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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}

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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.

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But this is no explanation why the symmetric matrices have the specified $\dim$. – Christian Ivicevic Aug 23 '12 at 10:03
Do you mean $n^2$? – enzotib Aug 23 '12 at 10:04
Yeah, I didn't notice earlier that he asked for proofs of the dimensions too, and have edited. – Rijul Saini Aug 23 '12 at 10:07
@RijulSaini: Thanks, but enzotib's answer seems to be easier to understand! – Christian Ivicevic Aug 23 '12 at 10:13
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