Prove that if an \(n \times n\) matrix X over R satisfies \(X ^ { 2 } = - I\) , then n is even.
Solution
1. One has\[ (\det X)^2=(-1)^n. \]
By multiplicativity of the determinant and the relation \(X^2=-I\), \[ (\det X)^2 = \det(X^2) = \det(-I) = (-1)^n. \]
2. One has \((-1)^n=1\).
Because \(X\) is a real matrix, \(\det X\in\RR\), so \[ (\det X)^2\geq0. \] By step 1, this square equals \((-1)^n\), which is either \(1\) or \(-1\). It therefore must equal \(1\).
3. The integer \(n\) is even.
By step 2, \[ (-1)^n=1. \] This holds exactly when \(n\) is even.
4. Q.E.D.
Step 3 is the required conclusion.