# Upper and lower bounds of eigenvalues for a 2 by 2 matrix

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?

• You can calculate the Gershgorin disc's radii for an estimate of where the eigenvalues reside. – mathreadler Jun 16 '17 at 14:52
• Welcome to the site by the way, you are encouraged to learn MathJax (LaTeX math) typesetting as it is used on the site also show any own attempts at solving a question especially if looks like could be a homework problem. – mathreadler Jun 16 '17 at 14:54
• For the bounds, do you have uncertainty in the entries of the matrix? Or is it just before computing? And do you need bounds on both, or one or the other? – Paul Jun 16 '17 at 14:57
• @Paul Thank you. I don't have uncertainty in the entries of the matrix and I need to know bounds on both. – Iman Hajizadeh Jun 16 '17 at 15:12
• @mathreadler Thank you. I need to know the bounds for eigenvalues because I need them in an optimization problem where I want to guarantee the stability of the matrix. I want to obtain this matrix in a way that the norm of eigenvalues is less than one. The eigenvalues can be real or complex numbers depending on the values of the matrix entries . – Iman Hajizadeh Jun 16 '17 at 15:20

Well, it is not hard to work it out explicitly, but observe what you get from working out the eigenvalues... Let $$\lambda_1$$ and $$\lambda_2$$ be the "largest" and "smallest" eigenvalues respectively (note $$\lambda_1$$ and $$\lambda_2$$ can be the complex conjugates of each other.

Denote $$\Delta=(a+d)^2 - 4(ad-bc) = (a-d)^2 - 4bc$$, we will call $$\Delta$$ the discriminant''.
Observe that the solutions to $$p(\lambda)= \lambda^2 - (a+d)\lambda + (ad-bc)=0$$ are $$\lambda = \frac{a+d}{2} \pm \frac{\sqrt{(a-d)^2 - 4bc}}{2}$$ (We now have everything we need to work out $$\lambda_1$$ and $$\lambda_2$$, but let's assume we didn't do the work, and/or have missing (or are given a range) for one of the elements in the matrix)

The above tells us

• the smallest value for $$\text{Real}(\lambda_1) = \frac{a+d}{2}$$
• the largest possible value for $$\text{Real}(\lambda_2) = \frac{a+d}{2}$$
• if $$(a-d)^2 - 4bc < 0$$, we have complex eigenvalues $$\frac{a+d}{2} \pm i\frac{\sqrt{\Delta}}{2}$$
• if $$(a-d)^2 - 4bc = 0$$, we have real repeated eigenvalues $$\frac{a+d}{2}$$
• if $$(a-d)^2 - 4bc > 0$$, we can use the series expansion for $$\sqrt{x^2 - 4y} \approx x - \frac{2y}{x}-\frac{2y^2}{x^3} -\frac{4y^3}{x^5} + O(\frac{1}{x^7})$$ with $$x=a-d$$ and $$y=bc$$ to give an upper bound on $$\lambda_1$$ and a lower bound on $$\lambda_2$$ (this is converges quickly if $$(a-d)^2>|bc|$$, (use Mclaurin series $$\sqrt{x-4y} \approx 2\sqrt{-y}+\frac{x^2}{\sqrt{-y}}+ O(x^4)$$ if $$(a-d)^2<|bc|$$ )

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 $$\lambda_{1} = \frac{1}{2} \xi = \frac{1}{2} \left( a + d \right)$$