Let A be a 2 \times 2 matrix with rational elements, and suppose there exists a natural number n such that A^n=−I, where I is the identity matrix of size 2. Prove that either A^2 = -I or A^3 =−I.

Given that \( A \) is a \( 2 \times 2 \) matrix with rational elements and that there exists a natural number \( n \) such that \( A^n = -I \), where \( I \) is the identity matrix of size 2, we need to prove that either \( A^2 = -I \) or \( A^3 = -I \).

### **Step-by-Step Proof:**

#### **Step 1: Consider the minimal polynomial of \( A \)**
Since \( A \) is a \( 2 \times 2 \) matrix, its minimal polynomial \( m_A(x) \) has a degree at most 2. The minimal polynomial is a monic polynomial of least degree such that \( m_A(A) = 0 \).

Also, \( A^n = -I \) implies that \( A \) satisfies the polynomial equation \( x^n + 1 = 0 \), meaning that \( A \) is a root of the polynomial \( x^n + 1 \). Therefore, the minimal polynomial \( m_A(x) \) divides \( x^n + 1 \).

#### **Step 2: Factorization of \( x^n + 1 \)**
The polynomial \( x^n + 1 \) factors as:
\[
x^n + 1 = (x + 1)(x^{n-1} - x^{n-2} + \cdots + x - 1)
\]
For example:
- If \( n = 2 \), \( x^2 + 1 \) factors as \( (x - i)(x + i) \).
- If \( n = 3 \), \( x^3 + 1 \) factors as \( (x + 1)(x^2 - x + 1) \).

Since \( A \) has rational elements, its minimal polynomial must have rational coefficients. Therefore, \( m_A(x) \) must be a factor of \( x^n + 1 \) that has rational coefficients.

The irreducible factors of \( x^n + 1 \) with rational coefficients can only be either \( x + 1 \) or factors of degree 2.

#### **Step 3: Possible minimal polynomials**
Given that \( A \) is a \( 2 \times 2 \) matrix, the minimal polynomial must have degree at most 2. The possible irreducible factors of \( x^n + 1 \) with degree at most 2 over the rationals are:
1. \( x^2 + 1 \)
2. \( x + 1 \) (but this is not applicable since it would imply \( A = -I \), contradicting the existence of \( n > 1 \))

If the minimal polynomial of \( A \) is \( x^2 + 1 \), then \( A \) satisfies the equation:
\[
A^2 + I = 0
\]
which implies:
\[
A^2 = -I
\]
Thus, if \( A^2 = -I \), the proof is complete.

#### **Step 4: Other possibilities**
Now, suppose that \( A^2 \neq -I \). This implies that the minimal polynomial of \( A \) is not \( x^2 + 1 \), but rather a factor of \( x^n + 1 \) for some \( n \). The next possible case is \( n = 3 \), where the minimal polynomial could be \( x^3 + 1 \). If this is the minimal polynomial, then \( A \) satisfies:
\[
A^3 + I = 0
\]
which implies:
\[
A^3 = -I
\]

Thus, if \( A^2 \neq -I \), then it must be the case that \( A^3 = -I \).

#### **Conclusion**
From the reasoning above, we conclude that either \( A^2 = -I \) or \( A^3 = -I \).

### Do you want details or have any questions?

### Further Questions:
1. What are the possible forms of the matrix \( A \) if \( A^2 = -I \)?
2. How does the minimal polynomial help in understanding the behavior of matrices?
3. Can this result be extended to higher-dimensional matrices?
4. What are the implications of this result for matrices with complex entries?
5. How can we explicitly construct a matrix \( A \) such that \( A^3 = -I \)?

**Tip:** The minimal polynomial is a key tool for analyzing matrix powers, and it can help identify the fundamental properties of a matrix, such as whether it is diagonalizable.

Learn how to prove that for a 2x2 matrix A with rational elements, there exists a natural number n such that A^n = -I. This problem involves understanding matrix algebra, minimal polynomials, and their implications on matrix powers. Suitable for advanced high school students and beyond.

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