Question

If a real matrix \(A\) satisfies \[ A^T=A \quad \text{and} \quad A^2=I, \] then the eigenvalues of \(A\) must be:

• A. Only \(1\)
• B. Only \(-1\)
• C. Either \(1\) or \(-1\)
• D. Zero only

Hint:

Whenever a matrix equation like \[ A^2=I \] appears, immediately convert it into an eigenvalue equation: \[ \lambda^2=1 \] This simplifies the problem instantly.

Answer 1

The Correct Option is C

Step 1 : Understanding the Question:
This linear algebra problem concerns finding the possible eigenvalues of a real matrix \( A \) that satisfies two properties: it is symmetric (\( A^T = A \)) and it is involutory (\( A^2 = I \)).
Step 2 : Key Formulas and Approach:
The definition of an eigenvalue \( \lambda \) and its corresponding non-zero eigenvector \( \mathbf{x} \) is:
\[ A\mathbf{x} = \lambda \mathbf{x} \] If we apply the matrix \( A \) repeatedly, we find that the eigenvalues of \( A^n \) are \( \lambda^n \):
\[ A^n\mathbf{x} = \lambda^n\mathbf{x} \] We will substitute the relation \( A^2 = I \) into this equation to constrain the values of \( \lambda \).
Step 3 : Detailed Solution:

Let \( \mathbf{x} \) be a non-zero eigenvector of \( A \) corresponding to the eigenvalue \( \lambda \).

Write the basic eigenvalue equation: \( A\mathbf{x} = \lambda \mathbf{x} \).

Multiply both sides on the left by matrix \( A \):
\[ A(A\mathbf{x}) = A(\lambda \mathbf{x}) \implies A^2\mathbf{x} = \lambda(A\mathbf{x}) \]
Substitute \( A\mathbf{x} = \lambda \mathbf{x} \) into the right side:
\[ A^2\mathbf{x} = \lambda^2 \mathbf{x} \]
Use the given condition \( A^2 = I \) on the left side:
\[ I\mathbf{x} = \lambda^2 \mathbf{x} \implies \mathbf{x} = \lambda^2 \mathbf{x} \]
Since the eigenvector \( \mathbf{x} \) is non-zero, we must have:
\[ \lambda^2 = 1 \implies \lambda = \pm 1 \]
Step 4 : Final Answer:
The eigenvalues of \( A \) must be either \( 1 \) or \( -1 \), which corresponds to option (C).
\[ \boxed{1 \text{ or } -1} \]