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 Incoming links: Orthogonal rotations generated by anticommuting involutions

Use

J E_{1} = - E_{2}

E_{1} J = E_{2}

and

J E_{1} J = E_{1}

to expand the matrix product:

A E_{1} A^{T} = c^{2} E_{1} + cs (J E_{1} - E_{1} J) - s^{2} J E_{1} J = (c^{2} - s^{2}) E_{1} - 2 cs E_{2} .

The double-angle identities therefore give

A E_{1} A^{T} = cos θ E_{1} - sin θ E_{2} .

Similarly

J E_{2} = E_{1}

E_{2} J = - E_{1}

and

J E_{2} J = E_{2}

, so

A E_{2} A^{T} = sin θ E_{1} + cos θ E_{2} .

This conjugation rotates the two-dimensional span of the two matrices through the double angle described by orthogonal rotations generated by anticommuting involutions.

The given symmetric matrices satisfy

E_{j}^{T} E_{j} = E_{j}^{2} = I

, so both are orthogonal matrices. Put

J = E_{1} E_{2}

c = cos (θ /2)

and

s = sin (θ /2)

. Anticommutation gives

J^{T} = E_{2} E_{1} = - J, J^{2} = E_{1} E_{2} E_{1} E_{2} = - I .

Thus

A = c I + s J

and

A^{T} = c I - s J

A A^{T} = (c^{2} + s^{2}) I = I

A - A^{T} = 2 s J

. Since

J^{2} = - I

makes

J

invertible, $A$ is symmetric exactly when $sin (θ /2) = 0$ .