Question

find all (n \times n) matrices (a) such that (a^2 = i), where (i) is the identity matrix.

Explanation:

🆕 New Concept Discovered: Involutory Matrices
Matrices that act as their own inverse.

Step 1: Define the matrix property

An \( n \times n \) matrix \( A \) satisfying \( A^2 = I \) is called an involutory matrix. This equation means that the matrix is its own multiplicative inverse:

$$ A = A^{-1} $$

Step 2: Analyze via diagonalization

The polynomial \( p(x) = x^2 - 1 = (x - 1)(x + 1) \) annihilates the matrix \( A \). Since this polynomial has distinct linear factors, the minimal polynomial of \( A \) must divide \( (x - 1)(x + 1) \).

Because the minimal polynomial splits into distinct linear factors, any such matrix \( A \) is diagonalizable.

Step 3: Determine the eigenvalues

The eigenvalues of \( A \) must be roots of the minimal polynomial. Therefore, the only possible eigenvalues of \( A \) are:

$$ \lambda = 1 \quad \text{or} \quad \lambda = -1 $$

Step 4: Characterize the general solution

Since \( A \) is diagonalizable, it can be written in the form:

$$ A = P D P^{-1} $$

where:

• \( P \) is any invertible \( n \times n \) matrix.
• \( D \) is a diagonal matrix whose diagonal entries are the eigenvalues of \( A \) (each being either \( 1 \) or \( -1 \)).

Thus, \( D \) takes the form:

$$ D =
LATEXBLOCK0
$$

for some integer \( p \) where \( 0 \le p \le n \), and where \( I_p \) is the \( p \times p \) identity matrix.

Answer:

The set of all such matrices \( A \) consists of all matrices of the form:

$$ A = P D P^{-1} $$

where \( P \) is any invertible \( n \times n \) matrix, and \( D \) is a diagonal matrix with diagonal entries in \(\{1, -1\}\).

Equivalently, these are the matrices that represent reflections across a subspace parallel to another subspace, covering all matrices of the form \( A = 2P - I \) where \( P \) is a projection matrix (\( P^2 = P \)).