Discussion
The matrix \(\begin{bmatrix} 1 & a \\ 8 & 3 \end{bmatrix}\) (where \(a \gt 0\)) has a negative eigenvalue if \(a\) is greater than
Answer: A
The matrix has a negative eigenvalue when a is greater than \(\dfrac{3}{8}\).
For a 2 x 2 matrix, the eigenvalues multiply to the determinant and add to the trace:
\[ \begin{aligned} \lambda_{1} + \lambda_{2} &= 1 + 3 = 4 \\ \lambda_{1}\lambda_{2} &= 3 - 8a \end{aligned} \]
The eigenvalues are real here, since \((1 - 3)^{2} + 4 \times 8a = 4 + 32a \gt 0\). With a positive sum, one eigenvalue is negative exactly when the product is negative: \(3 - 8a \lt 0\), that is \(a \gt 3/8\).
For a = 1/4 or 1/5 the determinant is 1 or 1.4, still positive, so both eigenvalues are positive. 1/8 comes from setting the determinant to \(1 - 8a\), forgetting the factor 3.
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