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:


But now how would one simplify this into the conditions above?


Note that by the formula you derived one of the conditions implies the other. So suppose $\lambda_1\geq \lambda_2$. Now you know that

$$ \begin{pmatrix} a-\lambda_1 & b\\ c & d-\lambda 1 \end{pmatrix} $$

is singular, i.e. there is a real number $r$ s.t.

$$ \begin{pmatrix} a-\lambda_1 \\ c \end{pmatrix} =r\begin{pmatrix} b\\ d-\lambda_1 \end{pmatrix} . $$

the case $a=d$ is easy because then $2\lambda_1\geq \lambda_1+\lambda_2=2a$ which implies your claim.

Now suppose that $a > d$ and $a>\lambda_1\geq d$. It is enough to consider this case, since $a>d\geq \lambda_1$ would imply that $\lambda_2\geq a>d$.

Now you know that $a-\lambda_1=rb$ and therefore $b\neq 0$. So you conclude that $r=(a-\lambda_1)/b$. But now this is a positive number and in order to have $c=r(d-\lambda_1)$ you must conclude that $c=d-\lambda_1=0$ and therefore $d=\lambda_1$. But this contradicts $\lambda_1 \geq \lambda_2$.


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.

  • $\begingroup$ Hi, how have you derived that equation? $\endgroup$ – user2850514 Apr 15 '16 at 10:42
  • $\begingroup$ It is well known "Courant-Fisher Theorem." $\endgroup$ – Sry Apr 18 '16 at 4:07

By Vieta's formula $\lambda_1 + \lambda_2 = a + d = S$ and $bc = ad - \lambda_1\lambda_2$. Suppose that $ a < d$ and $\lambda_1 < \lambda_2$. Function $ f(x) = x(S-x)$ is increasing on $(-\infty, \frac S 2)$. If $\lambda_1 > a$ then $\lambda_1\lambda_2 = f(\lambda_1) > f(a) = ad$. Therefore $bc < 0$. Contradiction.

  • $\begingroup$ How does this lead to the above conclusion? $\endgroup$ – user2850514 Apr 14 '16 at 20:16
  • $\begingroup$ Further conclusion is just not true. Put $a = \frac{9}{10}$, $d = \frac{1}{10}$, $\lambda_1 = \lambda_2 = \frac{1}{2}$ then $a + d = 1 = \lambda_1 + \lambda_2$ and you can choose positive $b$ and $c$ such that $bc = ad - \lambda_1\lambda_2$ $\endgroup$ – Maciej Starostka Apr 14 '16 at 20:53
  • $\begingroup$ @BigBolzano You actually can't choose $b,c$ positive with $bc=ad-\lambda_1\lambda_2$ since $ad-\lambda_1\lambda_2<0$. $\endgroup$ – Maik Pickl Apr 14 '16 at 21:47
  • $\begingroup$ You are right i edited $\endgroup$ – Maciej Starostka Apr 15 '16 at 6:48

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