91Ó°ÊÓ

For the Pauli spin matrix Ain Problem 6 , find the matricessin(kA) ,cos(kA) , ${\mathbf{e}}^{\mathbf{kA}}\mathbf{,}{\mathbf{e}}^{\mathbf{ikA}}\mathbf{,}$where $\mathbf{=}\sqrt{\mathbf{-}\mathbf{1}}$.

To see a physical example of non-commuting rotations, do the following experiment. Put a book on your desk and imagine a set of rectangular axes with the xand yaxes in the plane of the desk with the zaxis vertical. Place the book in the first quadrant with the x and yaxes along the edges of the book. Rotate the book$\mathbf{90}\mathbf{°}$about the xaxis and then$\mathbf{90}\mathbf{°}$about theaxis; note its position. Now repeat the experiment, this time rotating$\mathbf{90}\mathbf{°}$about theaxis first, and then$\mathbf{90}\mathbf{°}$about the xaxis; note the different result. Write the matrices representing the$\mathbf{90}\mathbf{°}$rotations and multiply them in both orders. In each case, find the axis and angle of rotation.

For each of the following matrices, find its determinant to see whether it produces a rotation or a reflection. If a rotation, find the axis and angle of rotation. If a reflection, find the reflecting plane and the rotation (if any) about the normal to this plane.

A particle is traveling along the line (x-3)/2=(y+1)/(-2)=z-1. Write the equation of its path in the form $\mathbf{r}\mathbf{=}{\mathbf{r}}_{\mathbf{0}}\mathbf{+}\mathbf{At}$. Find the distance of closest approach of the particle to the origin (that is, the distance from the origin to the line). If t represents time, show that the time of closest approach is $\mathbf{t}\mathbf{=}\mathbf{-}\left(r_0×A\right)\mathbf{/}{\left|A\right|}^{\mathbf{2}}$. Use this value to check your answer for the distance of closest approach. Hint: See Figure 5.3. If P is the point of closest approach, what is $\mathbf{A}\mathbf{×}{\mathbf{r}}^{\mathbf{2}}$?

Use vectors to prove the following theorems from geometry:

The line segment joining the midpoints of two sides of any triangle is parallel to the third side and half its length.

Show that the trace of a rotation matrix equals $\mathbf{2}\mathit{c}\mathit{o}\mathit{s}\mathit{θ}\mathbf{+}\mathbf{1}$ where θ is the rotation angle, and the trace of a reflection matrix equals $\mathbf{2}\mathit{c}\mathit{o}\mathit{s}\mathit{θ}\mathbf{-}\mathbf{1}$. Hint: See equations (7.18) and (7.19), and Problem 10.

Show that each of the following matrices is orthogonal and find the rotation and/or reflection it produces as an operator acting on vectors. If a rotation, find the axis and angle; if a reflection, find the reflecting plane and the rotation, if any, about the normal to that plane.

$\mathit{M}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\left(}\begin{array}{ccc}\mathbf{1}& \sqrt{\mathbf{2}}& \mathbf{-}\mathbf{1}\\ \sqrt{\mathbf{2}}& \mathbf{0}& \sqrt{\mathbf{2}}\\ \mathbf{1}& \mathbf{-}\sqrt{\mathbf{2}}& \mathbf{-}\mathbf{1}\end{array}\mathbf{\right)}$

In a parallelogram, the two lines from one corner to the midpoints of the two opposite sides trisect the diagonal they cross.

The Caley-Hamilton theorem states that "A matrix satisfies its own characteristic equation." Verify this theorem for the matrix $\mathit{M}$in equation (11.1). Hint: Substitute the matrix$\mathit{M}$forrole="math" localid="1658822242352" $\mathit{λ}$in the characteristic equation (11.4) and verify that you have a correct matrix equation. Further hint: Don't do all the arithmetic. Use (11.36) to write the left side of your equation as$\mathit{C}\left(D^2-7D+6\right){\mathit{C}}^{\mathbf{-}\mathbf{1}}$and show that the parenthesis$\mathbf{=}\mathbf{0}$. Remember that, by definition, the eigenvalues satisfy the characteristic equation.

Are the following linear vector functions? Prove your conclusions using (7.2).

4.$\mathit{F}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{=}\mathit{r}\mathbf{+}\mathit{A}$,whereAis a given vector.

Prove the following by appropriate manipulations using Facts 1 to 4; do not just evaluate the determinants.

$\left|\begin{array}{ccc}1& a& bc\\ 1& b& ac\\ 1& c& ab\end{array}\right|\mathbf{=}\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|\mathbf{=}\left(c-a\right)\left(b-a\right)\left(c-b\right)\left|\begin{array}{ccc}1& a& a^2\\ 0& 1& b+a\\ 0& 0& 1\end{array}\right|\mathbf{=}\left(c-a\right)\left(b-a\right)\left(c-b\right)$