Let $$A = \begin{bmatrix} \cos\theta & 0 & -\sin\theta \\ 0 & 1 & 0 \\ \sin\theta & 0 & \cos\theta \end{bmatrix}$$. If for some $$\theta \in (0, \pi)$$, $$A^2 = A^T$$, then the sum of the diagonal elements of the matrix $$(A + I)^3 + (A - I)^3 - 6A$$ is equal to ________.

Write $$A$$ in compact form by denoting $$c=\cos\theta$$ and $$s=\sin\theta$$:

$$A=\begin{bmatrix} c & 0 & -s \\ 0 & 1 & 0 \\ s & 0 & c \end{bmatrix}$$

First compute $$A^T$$ and $$A^2$$.

Transpose:
$$A^T=\begin{bmatrix} c & 0 & s \\ 0 & 1 & 0 \\ -s & 0 & c \end{bmatrix}$$

Square of $$A$$:
$$A^2=\begin{bmatrix} c & 0 & -s \\ 0 & 1 & 0 \\ s & 0 & c \end{bmatrix} \begin{bmatrix} c & 0 & -s \\ 0 & 1 & 0 \\ s & 0 & c \end{bmatrix} =\begin{bmatrix} c^2-s^2 & 0 & -2cs \\ 0 & 1 & 0 \\ 2cs & 0 & c^2-s^2 \end{bmatrix}$$

Using the double-angle identities, rewrite

$$A^2=\begin{bmatrix} \cos 2\theta & 0 & -\sin 2\theta \\ 0 & 1 & 0 \\ \sin 2\theta & 0 & \cos 2\theta \end{bmatrix}$$

The condition $$A^2=A^T$$ gives three independent equations:

$$\cos 2\theta = \cos\theta$$
$$-\sin 2\theta = \sin\theta$$
$$\sin 2\theta = -\sin\theta$$

Because $$\theta\in(0,\pi)$$, $$\sin\theta\neq0$$. Divide the second equation by $$\sin\theta$$:

$$-2\cos\theta = 1 \;\;\Longrightarrow\;\; \cos\theta = -\frac12$$

Thus $$\theta=\frac{2\pi}{3}$$, giving

$$c=-\frac12,\qquad s=\frac{\sqrt3}{2}$$

Next define the required polynomial in $$A$$:

$$E = (A+I)^3 + (A-I)^3 - 6A$$

Expand the two cubes separately:

$$(A+I)^3 = A^3 + 3A^2 + 3A + I$$
$$(A-I)^3 = A^3 - 3A^2 + 3A - I$$

Add them and subtract $$6A$$:

$$E = \big(A^3 + 3A^2 + 3A + I\big) + \big(A^3 - 3A^2 + 3A - I\big) - 6A$$
$$\;\; = 2A^3 + 6A - 6A = 2A^3$$

Therefore $$E=2A^3$$ and the trace of $$E$$ is simply twice the trace of $$A^3$$.

To find $$\operatorname{tr}(A^3)$$, use the eigenvalues of $$A$$. A$$ is a rotation matrix about the $$y$$-axis through $$$$\theta=\frac{2\pi}{3}$$$$, so its eigenvalues are

$$$$\lambda_1$$ = 1,\qquad $$\lambda_2$$ = e^{i$$\theta$$},\qquad $$\lambda_3$$ = e^{-i$$\theta$$}$$

Hence

$$$$\lambda_2^3$$ = e^{3i$$\theta$$}=e^{i2$$\pi$$}=1,\qquad $$\lambda_3^3$$ = e^{-3i$$\theta$$}=1$$

Thus

$$\operatorname{tr}(A^3)=$$\lambda_1^3+\lambda_2^3+\lambda_3^3$$ = 1+1+1 = 3$$

Finally,

$$\operatorname{tr}(E)=2\,\operatorname{tr}(A^3)=2$$\times$$3=6$$

Hence the sum of the diagonal elements of $$(A + I)^3 + (A - I)^3 - 6A$$ equals $$6$$.