I have a 2 by 2 matrix $$ \mathbf{A} = \left[ \begin{array}{c} a & b \\ c & d \end{array} \right], \qquad a, b, c, d \in \mathbb{C} $$ that has two eigenvalues (that can be real numbers or complex numbers). Is there any way that without direct calculating the eigenvalues, the upper and lower bounds of eigenvalues are specified? I know that the eigenvalues for this matrix are roots of the following polynomial:
$$ p(x) = x^{2} - x \text{ trace}\mathbf{A} + \det \mathbf{A} = x^2-(a+d) x+ a d - b c $$
and the sum of eigenvalues (roots) is the trace $(a+d)$ and their product is the determinant $(a d - b c)$.
Can we also find the upper and lower bounds for the roots of this polynomial (eigenvalues) without their direct calculation?
Start with a full rank matrix $$ \mathbf{A} = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] $$
As noted in your question, we have the equations for the trace and determinant to guide the solution.
Using the shorthand $$ \begin{align} \xi &= \text{trace} \mathbf{A} = a + d\\ \eta &= \det \mathbf{A} = ad - bc \end{align} $$ and denoting the eigenvalues by $\lambda$ and using the condition $$ \lambda \ne ,0$$ we can express the eigenvalues as $$ \lambda_{1} + \lambda_{2} = \xi \tag{1} $$ $$ \lambda_{1} \lambda_{2} = \eta \tag{2} $$
From $(1)$, $$ \lambda_{2} = \xi - \lambda_{1}. $$ Substitution into $(2)$ yields $$ \lambda_{1} \left( \xi - \lambda_{1} \right) = \eta \tag{3} $$ Minimize $(3)$ to find $$ \lambda_{1} = \frac{1}{2} \xi = \frac{1}{2} \left( a + d \right) $$