Question

Let $A$ be an $n \times n$ matrix of real numbers such that $A^2 = I$, then the eigen values of $A$ are

• A. 0, 1
• B. 1, -1
• C. 1, 1
• D. -1, -1

Hint:

Any polynomial relationship satisfied by a matrix \( p(A) = 0 \) is also satisfied by its eigenvalues \( p(\lambda) = 0 \).
For \( A^2 - I = 0 \), we have \( \lambda^2 - 1 = 0 \implies \lambda = \pm 1 \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
An eigenvalue \( \lambda \) and its corresponding eigenvector \( \mathbf{v} \) (where \( \mathbf{v} \neq \mathbf{0} \)) of a square matrix $A$ satisfy the defining relation:
\[ A\mathbf{v} = \lambda\mathbf{v} \]
This relationship allows us to use the matrix's algebraic equations to find its possible eigenvalues.

Step 2: Detailed Explanation:
Let us use the given matrix equation \( A^2 = I \) to solve for the eigenvalues:
Apply the matrix $A$ to both sides of our eigenvalue equation:
\[ A(A\mathbf{v}) = A(\lambda\mathbf{v}) \]
Using the property of linearity, we can factor out the scalar \( \lambda \):
\[ A^2\mathbf{v} = \lambda(A\mathbf{v}) \]
Substitute \( A\mathbf{v} = \lambda\mathbf{v} \) on the right side:
\[ A^2\mathbf{v} = \lambda(\lambda\mathbf{v}) = \lambda^2\mathbf{v} \]
Since we are given \( A^2 = I \), we can substitute \( I \) on the left side:
\[ I\mathbf{v} = \lambda^2\mathbf{v} \]
Because \( I\mathbf{v} = \mathbf{v} \):
\[ \mathbf{v} = \lambda^2\mathbf{v} \]
Since \( \mathbf{v} \neq \mathbf{0} \), we can equate the scalar coefficients:
\[ \lambda^2 = 1 \]
Solving this quadratic equation gives the possible values for \( \lambda \):
\[ \lambda = \pm 1 \]
Therefore, any eigenvalue of an involuntary matrix (\( A^2 = I \)) must be either $1$ or $-1$.

Step 3: Final Answer:
The eigenvalues of $A$ are 1, -1.