Question #15955

The Routh Hurwitz criterion tells us that the eigenvalues of a 2x2 matrix A
with real entries have negative real parts if the determinant of A is positive and the
trace of A (the sum of the entries on the main diagonal) is negative. Prove this.
1

Expert's answer

2012-10-09T09:17:52-0400

For matrix


(abcd)\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)


given condition are


adbc>0,a+d<0ad - bc > 0, \qquad a + d < 0


If we try to find eigenvalues


(aλbcdλ)\left( \begin{array}{cc} a - \lambda & b \\ c & d - \lambda \end{array} \right)(aλ)(bλ)bc=0(a - \lambda)(b - \lambda) - bc = 0λ2Bλ+C=0,B=a+d,C=adbc\lambda^2 - B\lambda + C = 0, \quad B = a + d, C = ad - bcλ=B±B24C2\lambda = \frac{B \pm \sqrt{B^2 - 4C}}{2}


we will see that both of them are negative.

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