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Most generic eigenvalues of a \(3\times 3\) traceless matrix with zero determinant are given as (\(a\) is some constant) :

This question was previously asked in
MH SET 2024 Paper 2 Computer Science and Applications Question Paper (07-Apr-2024)
The correct answer is

\(a, -a, 0\)

For any square matrix, the sum of the eigenvalues equals the trace and the product of the eigenvalues equals the determinant.

Let the eigenvalues be \(\lambda_{1},\lambda_{2},\lambda_{3}\). Traceless means \(\lambda_{1}+\lambda_{2}+\lambda_{3}=0\), and zero determinant means \(\lambda_{1}\lambda_{2}\lambda_{3}=0\).

The product condition forces at least one eigenvalue to vanish, say \(\lambda_{3}=0\). The trace condition then gives \(\lambda_{1}+\lambda_{2}=0\), so \(\lambda_{2}=-\lambda_{1}\).

Writing \(\lambda_{1}=a\), the most general set is \(a, -a, 0\).

The set \(0, 0, 0\) is only the special case \(a=0\); \(0, 0, a\) has trace \(a\), which is non-zero unless \(a=0\); and \(a, -2a, a\) has determinant \(-2a^{3}\), non-zero unless \(a=0\).

Check: \(a+(-a)+0=0\) (trace) and \(a\times(-a)\times 0=0\) (determinant), so both conditions hold for every value of \(a\).

Hence, \(a, -a, 0\).

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