natural symmetry of the Lorenz equation

The Lorenz equation has a natural symmetry defined by

$$(x,y,z)\mapsto (-x,-y,z).$$(1)

To verify that (1) is a symmetry of an ordinary differential equation (Lorenz equation) there must exist a $3\times 3$ matrix which commutes with the differential equation. This can be easily verified by observing that the symmetry is associated with the matrix $R$ defined as

(2)

Let

$$\dot{\text{𝐱}}=f(\text{𝐱})=\left[\begin{array}{c}\hfill \sigma (y-x)\hfill \\ \hfill x(\tau -z)-y\hfill \\ \hfill xy-\beta z\hfill \end{array}\right]$$(3)

where $f(\text{𝐱})$ is the Lorenz equation and ${\text{𝐱}}^T=(x,y,z)$. We proceed by showing that $Rf(\text{𝐱})=f(R\text{𝐱})$. Looking at the left hand side

$Rf(\text{𝐱})$$=$

$=$$\left[\begin{array}{c}\hfill \sigma (x-y)\hfill \\ \hfill x(z-\tau )+y\hfill \\ \hfill xy-\beta z\hfill \end{array}\right]$

and now looking at the right hand side

$f(R\text{𝐱})$$=$

$=$$f(\left[\begin{array}{c}\hfill -x\hfill \\ \hfill -y\hfill \\ \hfill z\hfill \end{array}\right])$

$=$$\left[\begin{array}{c}\hfill \sigma (x-y)\hfill \\ \hfill x(z-\tau )+y\hfill \\ \hfill xy-\beta z\hfill \end{array}\right].$

Since the left hand side is equal to the right hand side then (1) is a symmetry of the Lorenz equation.