The Lorenz equation has a natural symmetry defined by \begin{equation} (x,y,z) \mapsto (-x,-y,z). \label{eq:sym} \end{equation}To verify that () is a symmetry of an ordinary differential equation (Lorenz equation) there must exist a $3\times3$ 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 \begin{equation} R = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}. \end{equation}Let \begin{equation} \dot{\textbf{x}} = f(\textbf{x}) = \begin{bmatrix} \sigma(y-x) \\ x(\tau - z) -y \\ xy - \beta z \end{bmatrix} \end{equation}where $f({x})$ is the Lorenz equation and ${x}^T = (x,y,z)$ . We proceed by showing that $Rf({x}) = f(R{x})$ . Looking at the left hand side \begin{eqnarray*} Rf(\textbf{x}) & = & \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \sigma(y-x) \\ x(\tau - z) -y \\ xy - \beta z \end{bmatrix}\\ & = & \begin{bmatrix} \sigma(x-y) \\ x(z - \tau ) + y \\ xy - \beta z \end{bmatrix} \end{eqnarray*}and now looking at the right hand side \begin{eqnarray*} f(R\textbf{x}) & = & f(\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix})\\ & = & f(\begin{bmatrix} -x \\ -y \\ z \end{bmatrix})\\ & = & \begin{bmatrix} \sigma(x-y) \\ x(z - \tau ) + y \\ xy - \beta z \end{bmatrix}. \end{eqnarray*}Since the left hand side is equal to the right hand side then () is a symmetry of the Lorenz equation.