Why has the space $\{X\in M(3,\mathbb{R}) : X+X^T=0\}$ dimension $3$ over $\mathbb{R}$

Question

How can we determine this space $\{X\in M(3,\mathbb{R}) : X+X^T=0\}$ is $3$ dimensional over $\mathbb{R}$. Here I can find a linearly independent set which has $3$ elements. So I know the dimension is $\geq 3$. Thanks!

Answer 1

Note that $X + X^T = 0$ means $X^T = -X$.

This means, for the entries in $X$:

• the entries on the diagonal are $0$.

• the entries above the diagonal determine those below the diagonal.

This should alow you to conclude.

Answer 2

Let $A=\left(\begin{array}{ccc}a & b & c \\ d & e & f \\ g & h & i \\ \end{array}\right)$ Then $A+A^{\mathsf{T}}=\left(\begin{array}{ccc}2 a & b+d & c+g \\ b+d & 2 e & f+h \\ c+g & f+h & 2i \\ \end{array}\right)=0$

So $a=e=i=0$ and $d=-b,g=-c, h=-f$ leaving three degrees of freedom.

Answer 3

A basis of $M_3(\mathbb{R})$ is given by the $9$ matrices $E_{i,j}$, which have a $1$ at place $(i,j)$ and $0$ everywhere else. Now the subspace of skew-symmetric matrices has dimension $3$. To see this, show that $S_1=E_{1,2}-E_{2,1},S_2=E_{1,3}-E_{3,1},S_3=E_{2,3}-E_{3,2}$ is a basis.

More generally, the subspace of skew-symmetric matrices of size $n$ forms a Lie algebra, namely $\mathfrak{so}(n)$ of dimension $\frac{n(n-1)}{2}$. So your subspace is $\mathfrak{so}(3)$.