Exercise 3.2.3

• 1.

  We have the eigenvalues of $A(t)$ as: ${\lambda }_1=1+t+\sqrt{1+{(1+t)}^2},{\lambda }_2=1+t-\sqrt{1+{(1+t)}^2}$

• 1.

  At $t=0$, $A(0)=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$, the eigenvalues are ${\lambda }_1=1+\sqrt{2},{\lambda }_2=1-\sqrt{2}$. The eigenvectors are: $x_1=\left[\begin{array}{c}\hfill 1+\sqrt{2}\hfill \\ \hfill 1\hfill \end{array}\right],x_2=\left[\begin{array}{c}\hfill 1-\sqrt{2}\hfill \\ \hfill 1\hfill \end{array}\right]$.

For $y^{T}A(0)=\lambda y^T$, we have the same eigenvectors. To satisfy $y^{T}x=1$, we set $y_1=\frac{1}{4+2\sqrt{2}}\left[\begin{array}{c}\hfill 1+\sqrt{2}\hfill \\ \hfill 1\hfill \end{array}\right],y_2=\frac{1}{4+2\sqrt{2}}\left[\begin{array}{c}\hfill 1-\sqrt{2}\hfill \\ \hfill 1\hfill \end{array}\right]$

$\frac{\mathit{d\lambda }}{\mathit{dt}}=\left[\begin{array}{c}\hfill 1+\frac{1+t}{\sqrt{{(1+t)}^2+1}}\hfill \\ \hfill 1-\frac{1+t}{\sqrt{{(1+t)}^2+1}}\hfill \end{array}\right]$. At $t=0$, we have $\frac{\mathit{d\lambda }}{\mathit{dt}}=\left[\begin{array}{c}\hfill 1+\frac{1}{\sqrt{2}}\hfill \\ \hfill 1-\frac{1}{\sqrt{2}}\hfill \end{array}\right]$

Now compute $\frac{\mathit{dA}}{\mathit{dt}}=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$, so we have

This agrees with $\frac{\mathit{d\lambda }}{\mathit{dt}}$

• 1.

  $A(t)-A(0)=t\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$, it has eigenvalues ${\lambda }_1=0,{\lambda }_2=2t$, which are both $\ge 0$ when $t\ge 0$. So the change matrix is semidefinite.

It’s easy to see that $$\lambda \text{\_}1(t)=1+t+\sqrt{1+{(1+t)}^2}\ge 1+\sqrt{2}=\lambda \text{\_}1(0)\ge 0\ge \lambda \text{\_}2(t)=1+t-\sqrt{1+{(1+t)}^2}\ge \lambda \text{\_}2(0)=1-\sqrt{2}\$$