If for the matrix, $$A = \begin{bmatrix} 1 & -\alpha \\ \alpha & \beta \end{bmatrix}$$, $$AA^T = I_2$$, then the value of $$\alpha^4 + \beta^4$$ is:

We have $$A = \begin{bmatrix} 1 & -\alpha \\ \alpha & \beta \end{bmatrix}$$ and $$AA^T = I_2$$.

Computing $$AA^T = \begin{bmatrix} 1 & -\alpha \\ \alpha & \beta \end{bmatrix}\begin{bmatrix} 1 & \alpha \\ -\alpha & \beta \end{bmatrix} = \begin{bmatrix} 1 + \alpha^2 & \alpha - \alpha\beta \\ \alpha - \alpha\beta & \alpha^2 + \beta^2 \end{bmatrix}$$.

Setting this equal to $$I_2$$, from the $$(1,1)$$ entry: $$1 + \alpha^2 = 1$$, so $$\alpha^2 = 0$$, giving $$\alpha = 0$$.

From the $$(2,2)$$ entry: $$\alpha^2 + \beta^2 = 1$$, so $$\beta^2 = 1$$, giving $$\beta = \pm 1$$.

Therefore, $$\alpha^4 + \beta^4 = 0 + 1 = 1$$.