Let $$A = I_2 - 2MM^T$$, where M is real matrix of order $$2 \times 1$$ such that the relation $$M^TM = I_1$$ holds. If $$\lambda$$ is a real number such that the relation $$AX = \lambda X$$ holds for some non-zero real matrix X of order $$2 \times 1$$, then the sum of squares of all possible values of $$\lambda$$ is equal to:

We have $$A = I_2 - 2MM^T$$, where $$M$$ is a $$2 \times 1$$ real matrix with $$M^TM = I_1 = [1]$$, i.e., $$M^TM = 1$$.

Note that $$A$$ is a Householder reflection matrix. We need to find all $$\lambda$$ such that $$AX = \lambda X$$ for some non-zero $$X$$.

Case 1: $$X = M$$

$$AM = (I_2 - 2MM^T)M = M - 2M(M^TM) = M - 2M = -M$$

So $$\lambda = -1$$ is an eigenvalue.

Case 2: $$X$$ perpendicular to $$M$$, i.e., $$M^TX = 0$$.

$$AX = (I_2 - 2MM^T)X = X - 2M(M^TX) = X - 0 = X$$

So $$\lambda = 1$$ is an eigenvalue.

The eigenvalues of $$A$$ are $$\lambda = 1$$ and $$\lambda = -1$$.

Sum of squares = $$1^2 + (-1)^2 = 2$$.

The answer is $$\boxed{2}$$.