Exercise 3.2.6

• (1)

  Let $A=D+\mathit{tS}$, at $t=0$, we have $A(0)=D$, and $\frac{\mathit{d\lambda }}{\mathit{dt}}=y^T(0)\frac{\mathit{dA}(0)}{\mathit{dt}}x(0)$.

For $A(0)=D$, we see its eigenvalues are ${\lambda }_i=i$, where $i=1,2,\dots ,n$. We can easily solve for eigenvectors $x_i=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill ⋮\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \\ \hfill ⋮\hfill \\ \hfill 0\hfill \end{array}\right]$ where the 1 is on the $i$th position for $i=1,2,\dots ,n$. And $y_i^T=\left[\begin{array}{ccccccc}\hfill 0\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 0\hfill \end{array}\right]$.

So for a given ${\lambda }_i=i$, we have $\frac{\mathit{d\lambda }}{\mathit{dt}}=\left[\begin{array}{ccccccc}\hfill 0\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 0\hfill \end{array}\right]S\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill ⋮\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \\ \hfill ⋮\hfill \\ \hfill 0\hfill \end{array}\right]=S_{\mathit{ii}}$

The derivatives at $t=0$ of the eigenvalues $\lambda (t)$ of $A$ is just the diagonal entries of $S$.

• (1)

  When $t$ is small and $t>0$, $A=D+\mathit{tS}$, the eigenvalues of $D$ are just $1,2,\dots ,n$, so the eigenvalues of $A$ interlace with eigenvalues of $D$.

• (1)

  For any $t>0$, We apply Weyl inequalities, i.e.  ${\lambda }_{\mathit{min}}(D+\mathit{tS})={\lambda }_n(D+\mathit{tS})\le {\lambda }_n(D)+{\lambda }_1(\mathit{tS})=1+t{\lambda }_1(S)$, The last step comes from $S=\mathit{Q\Lambda }{Q}^T$ because it’s symmetric positive definite. And ${\lambda }_{\mathit{min}}(D+\mathit{tS})={\lambda }_n(D+\mathit{tS})\ge {\lambda }_n(D)+{\lambda }_n(\mathit{tS})=1+t{\lambda }_n(S)$

Similarly we have ${\lambda }_{\mathit{max}}(D+\mathit{tS})={\lambda }_1(D+\mathit{tS})\ge {\lambda }_1(D)+{\lambda }_n(\mathit{tS})=n+t{\lambda }_n(S)$ and ${\lambda }_{\mathit{max}}(D+\mathit{tS})={\lambda }_1(D+\mathit{tS})\le {\lambda }_1(D)+{\lambda }_1(\mathit{tS})=n+t{\lambda }_1(S)$