Discussion
A set V of complex numbers forms a two-dimensional real vector space under the usual addition of complex numbers and multiplication by real numbers. Let \(T : V \to V\) be a linear transformation defined as \(T(z) = \bar{z}\). Eigenvalues of T are :
Answer: D
The eigenvalues are 1, −1.
As a real vector space, \(V\) has the basis \(\{1, i\}\). Complex conjugation leaves real numbers unchanged and reverses the sign of imaginary ones:
\[ T(1) = 1, \qquad T(i) = -i \]
So in this basis \(T\) is the diagonal matrix with entries 1 and −1, and those are its eigenvalues. Equivalently, \(T^{2} = I\), so any eigenvalue satisfies \(\lambda^{2} = 1\), and since \(T\) is neither \(I\) nor \(-I\) both signs occur.
The pair 1, 1 would make \(T\) the identity, 0, 0 would make it nilpotent (impossible since \(T^2 = I\)), and 2, −2 would need \(T^{2} = 4I\).
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