1. Given data

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

Also, $\lambda$ satisfies $AX=\lambda X$ for some non-zero real $2\times 1$ matrix $X$.

This means $\lambda$ is an eigenvalue of $A$.

2. Interpretation of $M^TM=1$

Let $M=\begin{bmatrix}a\\ b\end{bmatrix}.$ Then $M^TM=a^2+b^2=1.$ So $M$ is a unit vector in $\mathbb{R}^2$.

3. Find eigenvalues of $A$

We use two natural directions in $\mathbb{R}^2$:

(i) Along the vector $M$

Compute: $AM=(I_2-2MM^T)M= M-2M(M^TM).$ Since $M^TM=1$, $AM=M-2M=-M.$ So $M$ is an eigenvector with eigenvalue $\lambda=-1.$

(ii) Any vector orthogonal to $M$

Let $N$ be a non-zero vector such that $M^TN=0.$ Then $AN=(I_2-2MM^T)N=N-2M(M^TN)=N-0=N.$ So every vector perpendicular to $M$ is an eigenvector with eigenvalue $\lambda=1.$

Thus the only possible eigenvalues are $\lambda=1,-1.$

4. Sum of squares of all possible values of $\lambda$

The possible values are $1$ and $-1$. Therefore, $1^2+(-1)^2=1+1=2.$

5. Final answer

$\boxed{2}$

6. Comparison with stored answer

Stored correct answer = $2$.

Our derived answer also equals $2$, so it agrees.