The dimension of the vector space of all trace-zero symmetric matrices

Question

Find the dimension of the vector space of all symmetric matrices of order $n \times n$ (real entries) and trace equal to zero.

Answer 1

Hints:

== The subspace $\;S\;$ of $\,n\times n\;$ symmetric matrices over some field $\,\Bbb F\;$ has dimension equal to

$$1+2+3+\ldots+n=\frac{n(n+1)}2$$

== The map $\;tr. : S\to\Bbb F\;$ is a linear map (in fact, a linear functional)

== We know $\;\dim_{\Bbb F}\Bbb F=1\;$

Well, now use the dimension theorem.

Answer 2

Let's consider the case $n=5$: $$ \pmatrix{ \bullet&\bullet&\bullet&\bullet&\bullet\\ \color{red}\circ&\bullet&\bullet&\bullet&\bullet\\ \color{red}\circ&\color{red}\circ&\bullet&\bullet&\bullet\\ \color{red}\circ&\color{red}\circ&\color{red}\circ&\bullet&\bullet\\ \color{red}\circ&\color{red}\circ&\color{red}\circ&\color{red}\circ&\color{red}\circ\\ } $$ We can view the black entries as free variables. The values of the red entries can be inferred from the black ones. Do you know why? (Note that the entry at the bottom right corner is red.)

Answer 3

$$V_F=\{A\in M_M(F)\:A^T=A , Trace(A=0)\}=$$ forall $A\in V_F$ we have $trac(A)=0$ and $a_{ij}=a_{ji}$ such that $A=[a_{ij}]_{n \times n}$ $$A=\begin{pmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ a_{12} & a_{22} & a_2^2 & \cdots & a_2^n \\ \vdots & \vdots& \vdots & \ddots & \vdots \\ a_{1n} & a_{2n} & a_{3n} & \cdots & (-a_{11}-...-a_{n-1 n-1}) \end{pmatrix}=a_{11}\begin{pmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots& \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & -1 \end{pmatrix}+a_{22}\begin{pmatrix} 0 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots& \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & -1 \end{pmatrix}+...+a_{n-1n-1}\begin{pmatrix} 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots& \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & -1 \end{pmatrix}+a_{12}\begin{pmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots& \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 0 \end{pmatrix}+a_{13}\begin{pmatrix} 0 & 0 & 1 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots& \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \end{pmatrix}+....+a_{1n}\begin{pmatrix} 0 & 0 & 0 & \cdots & 1 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots& \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 0 \end{pmatrix}+........+a_{n-1n-1}\begin{pmatrix} 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots& \vdots & \ddots & \vdots \\0 & 0 & 0 & \cdots & 1 \\ 0 & 0 & 0 & \cdots & 0 \end{pmatrix}\Rightarrow$$$$V_{F}=\{\sum_{i=1}^{n-1}a_{i}E_{ii}+\sum_{i\gt j}c_jE_{ij}\}=<E_{11},E_{22},...,E_{n-1\times n-1},E_{12,....E_{1n},.....,E_{n-1\times n}}>=<X>$$ such that $ X =\{E_{11},E_{22},...,E_{n-1\times n-1},E_{12},....E_{1n},.....,E_{n-1\times n}\}$ then $X$ generate $V_F $ and easily we can prove $ X$ is linear independent totally X is basis for $V_F$ and $$|x|=\frac{(n-1)(n+2)}{2}$$