Key Concept: Eigenvalues and Properties of Special Matrices

This problem primarily revolves around the concept of eigenvalues and eigenvectors. For a given square matrix $A$, if $A X = \lambda X$ holds for some non-zero vector $X$ and scalar $\lambda$, then $\lambda$ is called an eigenvalue of $A$ and $X$ is its corresponding eigenvector. The problem also requires us to analyze the properties of the given matrix $A$ by calculating its square, $A^2$, which will reveal that $A$ is an involutory matrix. An involutory matrix is a square matrix that is its own inverse, meaning $A^2 = I$.

---

1. Understanding the Given Information

We are given:

• $A = I_2 - 2 M M^T$, where $I_2$ is the $2 \times 2$ identity matrix.
• $M$ is a real matrix of order $2 \times 1$ (a column vector). Let $M = \begin{pmatrix} m_1 \\ m_2 \end{pmatrix}$.
• The relation $M^T M = I_1$ holds. $I_1$ is the $1 \times 1$ identity matrix, which is simply the scalar $1$.
• $\lambda$ is a real number such that $A X = \lambda X$ for some non-zero real matrix $X$ of order $2 \times 1$ (an eigenvector).

Let's first interpret the condition $M^T M = I_1$. If $M = \begin{pmatrix} m_1 \\ m_2 \end{pmatrix}$, then $M^T = \begin{pmatrix} m_1 & m_2 \end{pmatrix}$. So, $M^T M = \begin{pmatrix} m_1 & m_2 \end{pmatrix} \begin{pmatrix} m_1 \\ m_2 \end{pmatrix} = [m_1^2 + m_2^2]$. The condition $M^T M = I_1$ means $[m_1^2 + m_2^2] = [1]$, which implies $m_1^2 + m_2^2 = 1$. This tells us that $M$ is a unit vector.

Now, let's look at the product $M M^T$: $M M^T = \begin{pmatrix} m_1 \\ m_2 \end{pmatrix} \begin{pmatrix} m_1 & m_2 \end{pmatrix} = \begin{pmatrix} m_1^2 & m_1 m_2 \\ m_2 m_1 & m_2^2 \end{pmatrix}$. This is a $2 \times 2$ matrix. It's important to note the difference in dimensions and type of matrix for $M^T M$ (a scalar) and $M M^T$ (a $2 \times 2$ matrix).

---

2. Step-by-Step Working

Step 2.1: Analyze the properties of $M M^T$ Let $P = M M^T$. We want to see how $P$ behaves when multiplied by itself. $P^2 = (M M^T)(M M^T)$ Using the associativity of matrix multiplication, we can group terms: $P^2 = M (M^T M) M^T$ We know from the given information that $M^T M = I_1 = [1]$. When a $1 \times 1$ matrix (a scalar) is multiplied with other matrices, it behaves like a scalar multiplication. $P^2 = M (1) M^T$ $P^2 = M M^T$ So, $P^2 = P$. This means the matrix $P = M M^T$ is an idempotent matrix. Specifically, it's a projection matrix onto the subspace spanned by $M$.

Step 2.2: Calculate $A^2$ The matrix $A$ is given by $A = I_2 - 2 M M^T$. We need to find $A^2$ to understand its fundamental properties. $A^2 = (I_2 - 2 M M^T)(I_2 - 2 M M^T)$ This is a standard algebraic expansion for matrices, similar to $(a-b)^2 = a^2 - 2ab + b^2$, but we must maintain matrix multiplication order. $A^2 = I_2 \cdot I_2 - I_2 (2 M M^T) - (2 M M^T) I_2 + (2 M M^T)(2 M M^T)$ Since $I_2$ is the identity matrix, $I_2 \cdot X = X \cdot I_2 = X$ for any matrix $X$ of compatible dimensions. $A^2 = I_2 - 2 M M^T - 2 M M^T + 4 (M M^T)(M M^T)$ $A^2 = I_2 - 4 M M^T + 4 (M M^T)^2$ Now, substitute the result from Step 2.1, $(M M^T)^2 = M M^T$: $A^2 = I_2 - 4 M M^T + 4 (M M^T)$ $A^2 = I_2$ This result is significant! It means $A$ is an involutory matrix.

Step 2.3: Find the possible values of $\lambda$ We are given the eigenvalue relation $A X = \lambda X$, where $X$ is a non-zero vector. To find $\lambda$, we can use the property $A^2 = I_2$. Multiply the eigenvalue equation by $A$ from the left: $A (A X) = A (\lambda X)$ $A^2 X = \lambda (A X)$ Now, substitute $A^2 = I_2$ on the left side: $I_2 X = \lambda (A X)$ And substitute $A X = \lambda X$ again on the right side: $X = \lambda (\lambda X)$ $X = \lambda^2 X$ Rearrange the equation: $X - \lambda^2 X = 0$ $(1 - \lambda^2) X = 0$ Since $X$ is a non-zero matrix (vector), for the product $(1 - \lambda^2) X$ to be zero, the scalar factor $(1 - \lambda^2)$ must be zero. $1 - \lambda^2 = 0$ $\lambda^2 = 1$ This gives us two possible values for $\lambda$: $\lambda = \pm 1$ So, the possible real values of $\lambda$ are $1$ and $-1$.

Step 2.4: Calculate the sum of squares of all possible values of $\lambda$ The possible values of $\lambda$ are $1$ and $-1$. The sum of squares of these values is: $(1)^2 + (-1)^2 = 1 + 1 = 2$

---

3. Tips for Success & Common Pitfalls

• Matrix Dimensions: Always pay attention to the dimensions of matrices. $M$ is $2 \times 1$, $M^T$ is $1 \times 2$. Thus, $M^T M$ is $1 \times 1$ (a scalar), while $M M^T$ is $2 \times 2$. This distinction is crucial.
• Matrix Multiplication Order: Matrix multiplication is not commutative ($AB \neq BA$ in general). Maintain the correct order of matrices when expanding products like $(I - 2P)^2$.
• Recognizing Special Matrices: Identifying that $M M^T$ is an idempotent matrix ($P^2=P$) simplifies the calculation of $A^2$ significantly. Similarly, recognizing $A^2=I$ means $A$ is an involutory matrix, which has direct implications for its eigenvalues.
• Eigenvalue Definition: The relation $AX = \lambda X$ is the fundamental definition. Understanding how to manipulate this equation by multiplying by $A$ again is a common technique to find relationships between $\lambda$ and properties of $A$.
• Non-zero Vector $X$: The condition that $X$ is a non-zero vector is critical for concluding that $1 - \lambda^2 = 0$. If $X$ could be the zero vector, then $X = \lambda^2 X$ would hold trivially for any $\lambda$.

---

4. Summary/Key Takeaway

The problem demonstrates how properties of a matrix ($A^2 = I_2$) directly influence its eigenvalues. By carefully calculating $A^2$ and utilizing the given condition $M^T M = I_1$, we found that $A$ is an involutory matrix. For any involutory matrix $A$, its eigenvalues must satisfy $\lambda^2 = 1$, leading to $\lambda = \pm 1$. The sum of squares of these possible eigenvalues is $1^2 + (-1)^2 = 2$. This approach avoids directly solving the characteristic equation $\det(A - \lambda I) = 0$, which would be more complex.

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