Exercise 2.72 (1) Correction

[jcwang111]

Hi, I believe in the solution given for 2.72 (1), the 1/2 factor in a and d is forgotten. So I believe it should be:
If $\rho$ is positive, all eigenvalues of $\rho$ should be non-negative.
\begin{align*}
\det (\rho - \lambda I) &= (a- \lambda) (b - \lambda) - |b|^2 = \lambda^2 - (a+d)\lambda + ad - |b^2| = 0\
\lambda &= \frac{(a+d) \pm \sqrt{(a+d)^2 - 4 (ad - |b|^2)}}{2}\
&= \frac{1/2 \pm \sqrt{1/4 - 4 \left(\frac{1 - r_3^2}{4} - \frac{r_1^2 + r_2^2}{4} \right)}}{2}\
&= \frac{1/2 \pm \sqrt{1/4 - (1 - r_1^2 - r_2^2 - r_3^2)}}{2}\
&= \frac{1/2 \pm \sqrt{|\vec{r}|^2-3/4}}{2}\
.\end{align*}
Because $|\vec{r}|^2 \le 1$, $\sqrt{|\vec{r}|^2-3/4} \le \sqrt{1/4} = 1/2$ if real, and $\lambda$ is non-negative.