We are tasked to prove that $T^2 = -I$ under the conditions that $T \in L(V)$, $T^4 = I$, and $T$ has no eigenvalues.

Solution

Step 1: Characteristic properties of $T$

1. $T^4 = I$ implies that $T$ is diagonalizable over $\mathbb{C}$ because its minimal polynomial divides $x^4 - 1$, and $x^4 - 1 = (x^2 - 1)(x^2 + 1)$ has distinct roots.

   The eigenvalues of $T$ are among the fourth roots of unity: $\lambda \in \{ 1, -1, i, -i \}.$

2. $T$ has no eigenvalues in $\mathbb{C}$. Therefore, $\lambda = 1$ and $\lambda = -1$ are not eigenvalues of $T$, leaving $\lambda = i$ and $\lambda = -i$ as the only possibilities.

Step 2: Minimal polynomial of $T$

• Since $T^4 = I$, the minimal polynomial $m_T(x)$ divides $x^4 - 1$. The absence of $\lambda = 1$ and $\lambda = -1$ as eigenvalues implies $m_T(x)$ does not have $(x-1)$ or $(x+1)$ as factors. Thus, $m_T(x) = x^2 + 1$.

• The minimal polynomial $x^2 + 1$ implies: $T^2 + I = 0 \implies T^2 = -I.$

Step 3: Verifying consistency

• Since $T^2 = -I$, we calculate $T^4$ as: $T^4 = (T^2)^2 = (-I)^2 = I,$ which is consistent with the given $T^4 = I$.

Conclusion

We have shown that $T^2 = -I$ based on the absence of eigenvalues for $T$, its diagonalizability, and the properties of the minimal polynomial.

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Would you like further clarification or details on any step?

Related Questions

1. How is the minimal polynomial determined for a linear operator?
2. What does it mean for a linear operator to have no eigenvalues?
3. Why does $T^4 = I$ imply $T$ is diagonalizable?
4. What are the implications of $T^2 = -I$ in terms of geometry or symmetry?
5. How would the proof change if eigenvalues $\lambda = 1$ or $\lambda = -1$ were allowed?

Tip:

When working with minimal polynomials, always verify divisibility by the characteristic polynomial to confirm compatibility with eigenvalue constraints.