Show that all eigenvalues of A have nonzero real part if T*D /= 0
Show that all eigenvalues of A have real part < 0 if D > 0 and T < 0
Based on the characteristic polynomial of A, I know that
T= a1 + a2, where a1 and a2 are eigenvalues of A
However, I do not know how to proceed from here. Any help would be greatly appreciated!!
Dec 5th 2009, 10:43 AM
You use the quadratic formula to show that all eigenvalues of a have the form .
This will show you that if the trace is zero then you have zero real part if . However, if you look at for you should notice that this implies that .
Dec 5th 2009, 12:14 PM
Thank you! I need to do the same with the 3x3 matrix case, but the characteristic polynomial is a cubic instead of a quadratic.
Let A be a 3x3 matrix.
char poly(A) = z^3 -Tz^2 + Mz - D = 0
Show that all eigenvalues of A have nonzero real part if (TM - D)D /= 0
Show that all eigenvalues of A are < 0 if D < 0, T < 0, and TM < D
T = tr(A) , D = det(A), and there's not a given definition for M.
Again, if a1, a2, a3 are eigenvalues of A, I found that
T = a1 +a2 + a3
M = a1*a2 + a2*a3 + a1*a3
D = a1*a2*a3
TM - D = (a1 + a2)*(a1 + a3)*(a2 + a3)