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 $A^3 = -I$.

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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.