Finding conditions on the eigenvalues of a matrix Consider the $2\times2$ matrix 
$$A = \begin{pmatrix}a&b\\c&d\end{pmatrix}$$
where $a,b,c,d\ge 0$. Show that $\lambda_1\ge\max(a,d)>0$ and $\lambda_2\le\min(a,d)$.

So the eigenvalues are given by the characteristic polynomial
$$(a-\lambda)(d-\lambda)-bc=0\implies \lambda^2 - (a+d)\lambda + ad - bc=0$$
And so the solutions to this equation are
$$\lambda_{1,2} = \frac{(a+d)\pm\sqrt{(a+d)^2-4(ad-bc)}}2 = \frac{(a+d)\pm\sqrt{(a-d)^2+4bc}}2$$
We may therefore simplify this to:
$$\lambda_1+\lambda_2=a+d$$
But now how would one simplify this into the conditions above?

You can look at this way also:
For any nonzero vector $x$, we know that,
$$\lambda_1 \ge x^TAx/{x^Tx}$$ and $$\lambda_n \le x^TAx/{x^Tx}$$ where $\lambda_1\ge \lambda_2\ge \ldots\ge\lambda_n$ are eigenvalues of $A$. So, taking $x$ as $ \begin{pmatrix}1\\0\end{pmatrix}$ and $ \begin{pmatrix}0\\1\end{pmatrix}$, we get the desired result.
