Discussion
If A is a \(2 \times 2\) real matrix with trace zero and determinant 1, then the eigenvalues of A are :
Answer: D
The eigenvalues are purely imaginary.
For a \(2 \times 2\) matrix the characteristic polynomial is \(\lambda^{2} - (\text{trace})\lambda + \det\). With trace 0 and determinant 1:
\[ \lambda^{2} + 1 = 0 \quad \Rightarrow \quad \lambda = \pm i \]
An example is the rotation \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\).
Real eigenvalues would need the discriminant \(\text{tr}^{2} - 4\det = -4\) to be non-negative, which it is not, so they are neither real and distinct nor real and repeated. The real part of the eigenvalues is half the trace, which is 0, so 'complex with non-zero real part' is also ruled out.
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