
eigenvalues
Let A be a 2x2 matrix and let
char poly(A) = z^2  Tz + D = 0.
T=trace(A) D=det(A)
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
D=a1*a2
However, I do not know how to proceed from here. Any help would be greatly appreciated!!

You use the quadratic formula to show that all eigenvalues of a $\displaystyle 2 \times 2$ have the form
$\displaystyle z = \frac{\text{tr}(A) \pm \sqrt{\text{tr}(A)  4\det(A)}}{2}$.
This will show you that if the trace is zero then you have zero real part if $\displaystyle \det(A) \ge 0$. However, if you look at $\displaystyle z^2  \text{tr}(A) z = 0$ for $\displaystyle \det(A) = 0$ you should notice that this implies that $\displaystyle z = \{0, \text{tr(A)}\}$.

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)