1. (i) Let$$A=\left(\begin{array}{ccc}1 & a & b \\ c & d & -1 \\ e & \frac{1}{2} & f\end{array}\right)$$Are there constants $a, b, c, d, e, f$ such that $A$ is orthogonal?
   Solution.
   If $A$ is orthogonal, the modulus of every row and column is 1, therefore all other elements in the same row with ±1 must be 0, so $a=b=c=d=e=f=0$. The modulus of last row is not 1, so $A$ cannot be orthogonal.
   (ii) If an orthogonal matrix represents a reflection, show that it is symmetric. Is the converse true?
   Solution.
   For all $\mathbf{v},\mathbf{x}\in\mathbb R^{1×3}$, the vector of $\mathbf x$ projected onto $v$ is $\frac{\mathbf v^{\sf T}\mathbf x}{\sqrt{\mathbf v^{\sf T}\mathbf v}}\cdot\frac{\mathbf v}{\sqrt{\mathbf v^{\sf T}\mathbf v}}$. This vector is the midpoint of $\mathbf x$ and the reflection of $\bf x$ in the hyperplane containing $\bf v$, therefore the reflected vector is $Q\mathbf x$, where $Q=I-2{\frac {{\mathbf {v} }{\mathbf {v} }^{\sf T }}{{\mathbf {v} }^{\sf T }{\mathbf {v} }}}$. Since $I,\mathbf v\mathbf v^{\sf T}$ are symmetric, so is $Q$.[Another way: If $A$ is orthogonal, $A^2=I=A^{\sf T}A$, so $A=A^2A^{\sf T}=A^{\sf T}(AA^{\sf T})=A^{\sf T}$, so $A$ is symmetric.]
   The converse is not true. Counter-example: a diagonal matrix is symmetric. For it to be orthogonal, every diagonal element is ±1. Among them, only if there is one -1 and all the others 1, is it matrix of a reflection, so all the other cases can be taken as counter-example.
   (iii) Let $A$ be an $n × n$ matrix for which there exists an orthogonal matrix $P$ such that $P^{\sf T}AP$ is diagonal. Show that $A$ is symmetric.
   Proof.
   $P^{\sf T}AP$ is diagonal⇒$(P^{\sf T}AP)^{\sf T}=P^{\sf T}AP$⇒$P^{\sf T}A^{\sf T}P=P^{\sf T}AP$. Since $P$ is orthogonal, $P^{\sf T}=P^{-1}$, so $A^{\sf T}=A$.
2. (i) Let $T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}$ be the map given by$$T(\mathbf{x})=A \mathbf{x}+\mathbf{b} \text { where } A=\left(\begin{array}{cc}1 / 3 & 0 \\ 2 / 3 & 1 / \sqrt{2} \\ 2 / 3 & -1 / \sqrt{2}\end{array}\right) \quad \text { and } \mathbf{b}=\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right)$$Show that $A^{\sf T}A = I_2$ and deduce that $T$ is an isometry. What is the image of $T$?
   Proof.
   $A^{\sf T}A=\begin{pmatrix}\left(\frac13\right)^2+\left(\frac23\right)^2+\left(\frac23\right)^2&\frac13\cdot0+\frac23\cdot\frac1{\sqrt2}+\frac23\cdot\left(-\frac1{\sqrt2}\right)\\\frac13\cdot0+\frac23\cdot\frac1{\sqrt2}+\frac23\cdot\left(-\frac1{\sqrt2}\right)&0^2+\left(\frac1{\sqrt2}\right)^2+\left(-\frac1{\sqrt2}\right)^2\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$. So $T$ is an isometry.
   Let $B=(\begin{array}3-4&1&1\end{array})$, we have $BA=0$, so the image of $T$ is the plane $-4x+y+z=B\mathbf b$, which is $-4x+y+z=-2$.
   (ii) Show that there is no isometry from $\mathbb R^3$ to $\mathbb R^2$.
   Take an orthogonal basis of $\mathbb R^3$, since isometry preserves orthogonality, their image would be pairwise-orthogonal, which is impossible in $\mathbb R^2$.
   Another way: Take 4 vertices of a tetrahedron in $\mathbb R^3$, they are all equidistant, after mapping $T$ to $\mathbb R^2$, they cannot be equidistant.
   Assume such a $T$ exists, WLOG, $T(0,0,0)=(0,0)$ by choice of origin, $T(1,0,0)=(1,0)$ by rotation, $T(0,1,0)=(a,b)$. Now $|(1,0,0)-(0,1,0)|=\sqrt2$. If $T$ is isometry, we must have $(a-1)^2+b^2=2$, and $|(0,1,0)-(0,0,0)|=1$, hence $a^2+b^2=1$, solving, $a=0,b=\pm1$, $\therefore T(0,1,0)=(0,\pm1)$. Now let $T(0,0,1)=(c,d)$, by same argument as for $T(0,1,0)$, we get $(c,d)=(0,\pm1)$. Hence either $T(0,0,1)=T(0,1,0)$⇒distance=0, not isometry; or $T(0,0,1)=-T(0,1,0)$⇒distance=2, not isometry.
3. Let $\mathbf e_1,\mathbf  e_2,\mathbf  e_3$ be an orthonormal basis in $\mathbb R^3$ which is right-handed. Say that$$X \mathbf{e}_{1}+Y \mathbf{e}_{2}+Z \mathbf{e}_{3}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} \quad \text { and } \quad \tilde{X} \mathbf{e}_{1}+\tilde{Y} \mathbf{e}_{2}+\tilde{Z} \mathbf{e}_{3}=\tilde{x} \mathbf{i}+\tilde{y} \mathbf{j}+\tilde{z} \mathbf{k}$$Show that$$X \tilde{X}+Y \tilde{Y}+Z \tilde{Z}=x \tilde{x}+y \tilde{y}+z \tilde{z}$$and$$(Y \tilde{Z}-Z \tilde{Y}) \mathbf{e}_{1}+(Z \tilde{X}-X \tilde{Z}) \mathbf{e}_{2}+(X \tilde{Y}-Y \tilde{X}) \mathbf{e}_{3}=(y \tilde{z}-z \tilde{y}) \mathbf{i}+(z \tilde{x}-x \tilde{z}) \mathbf{j}+(x \tilde{y}-y \tilde{x}) \mathbf{k}$$What is the significance of these identities?
   Proof.
   Since $\mathbf e_1,\mathbf  e_2,\mathbf  e_3$ is an orthonormal basis, we have $\mathbf e_i\cdot\mathbf e_j=\delta_{ij}$ and $\mathbf e_i\wedge\mathbf e_i=0$, $\mathbf e_1\wedge\mathbf e_2=-\mathbf e_2\wedge\mathbf e_1=\mathbf e_3$, $\mathbf e_3\wedge\mathbf e_1=-\mathbf e_1\wedge\mathbf e_3=\mathbf e_2$.
   By distributivity, we have $X \tilde{X}+Y \tilde{Y}+Z \tilde{Z}=x \tilde{x}+y \tilde{y}+z \tilde{z}$ and $(Y \tilde{Z}-Z \tilde{Y}) \mathbf{e}_{1}+(Z \tilde{X}-X \tilde{Z}) \mathbf{e}_{2}+(X \tilde{Y}-Y \tilde{X}) \mathbf{e}_{3}=(y \tilde{z}-z \tilde{y}) \mathbf{i}+(z \tilde{x}-x \tilde{z}) \mathbf{j}+(x \tilde{y}-y \tilde{x}) \mathbf{k}$.
   Significance: Formula for dot and vector product is unchanged between orthonormal bases. After multivariable calculus, compare this with $\nabla\cdot,\nabla\wedge$, which do not have different expressions in cartesian vector
4. Let $\mathbf u,\mathbf v,$ and $\mathbf w$ be vectors in $\mathbb R^3$. What does it mean to say that $\{\mathbf u,\mathbf v,\mathbf w\}$ forms a basis for $\mathbb R^3$?
   (i) Suppose a vector $x ∈\mathbb R^3$ has coordinates $(x_1, x_2, x_3)$ in the basis $\{\mathbf u,\mathbf v,\mathbf w\}$. Give a formula for the length of $x$ in terms of its coordinates and properties of $\mathbf u,\mathbf v,\mathbf w$. What properties of $\mathbf u,\mathbf v,\mathbf w$ are thus required in order for the usual length formula to hold?
   Hence or otherwise, determine the equation of the unit sphere in the coordinate system defined by the basis $\{\mathbf u,\mathbf v,\mathbf w\}$.
   (ii) Suppose $\{\mathbf u,\mathbf v,\mathbf w\}$ and $\{\mathbf U,\mathbf V,\mathbf W\}$ are both orthonormal bases in $\mathbb R^3$, with coordinates given by $\mathbf x = (x_1, x_2, x_3)$ and $\mathbf X = (X_1, X_2, X_3)$, respectively. By considering the transformation of coordinates in converting from one basis to the other, show that the equation of the unit sphere is invariant.
   Solution.
   It means they span $\mathbb R^3$ and they are linearly independent.
   (i)$|\mathbf x|^2=x_1^2\mathbf u^2+x_2^2\mathbf v^2+x_3^2\mathbf w^2+2x_1x_2\mathbf u\cdot\mathbf v+2x_2x_3\mathbf v\cdot\mathbf w+2x_3x_1\mathbf w\cdot\mathbf u$.
   Usual formula: $|\mathbf x|=\sqrt{x_1^2+x_2^2+x_3^2}$. $\{\mathbf u,\mathbf v,\mathbf w\}$ are required to be orthonormal. Hence the unit sphere has equation $x_1^2+x_2^2+x_3^2=1$.
   (ii)Let the matrix of the transformation of coordinates in converting from one basis to the other be $A$, then the sphere has equation $\mathbf x^{\sf T}\mathbf x=1$, which transforms to $(A\mathbf x)^{\sf T}A\mathbf x=1$⇔$\mathbf x^{\sf T}A^{\sf T}A\mathbf x=1$. Because $A^{\sf T}A=I$, we have $\mathbf x^{\sf T}\mathbf x=1$, so the equation of the unit sphere is invariant.
5. Let $\mathbf n$ be a unit vector in $\mathbb R^3$. The stretch $S_k$ with invariant plane $\mathbf r ·\mathbf n = c$ and stretch factor $k$ is defined by $S_{k}(\mathbf{v})=\mathbf{v}+(k-1)(\mathbf{v} \cdot \mathbf{n}-c) \mathbf{n}$.
   (i) Describe the maps $S_1, S_0$ and $S_{−1}$.
   (ii) Determine the matrix for $S_k$ when the invariant plane is $x + y + z = 0$.
   (iii) Find an orthonormal basis $\mathbf e_1,\mathbf e_2,\mathbf e_3$ of $\mathbb R^3$ such that $\mathbf e_1,\mathbf e_2$ are parallel to the plane in (ii). Describe the map $S_k$ in terms of the coordinates associated with $\mathbf e_1,\mathbf e_2,\mathbf e_3$.
   Solution.
   (i) $S_1$ is identity transformation. $S_0$ is projection on the plane. $S_{-1}$ is reflection about the plane.
   (ii) $\mathbf{v}=\left(\begin{array}3x&y&z\end{array}\right)$,$S_{k}(\mathbf{v})=\left(\begin{array}3x&y&z\end{array}\right)+(k-1)(x+y+z)\left(\begin{array}3\frac1{\sqrt3}&\frac1{\sqrt3}&\frac1{\sqrt3}\end{array}\right)$, so the matrix is $\frac1{\sqrt3}\begin{pmatrix}k-1+\sqrt3&k-1&k-1\\k-1&k-1+\sqrt3&k-1\\k-1&k-1&k-1+\sqrt3\end{pmatrix}$.
   (iii) $\mathbf e_1=\left(\frac1{\sqrt2},-\frac1{\sqrt2},0\right)$,$\mathbf e_2=\left(\frac1{\sqrt2},0,-\frac1{\sqrt2}\right)$,$\mathbf e_3=\left(0,\frac1{\sqrt2},\frac1{\sqrt2}\right)$.
   $\left(
   \begin{array}{ccc}
   k & k-1 & k-1 \\
   k-1 & k & k-1 \\
   k-1 & k-1 & k \\
   \end{array}
   \right).\left(
   \begin{array}{ccc}
   \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\
   \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\
   0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
   \end{array}
   \right)^{\sf T}=\left(
   \begin{array}{ccc}
   \frac{k}{\sqrt{2}}-\frac{k-1}{\sqrt{2}} & \frac{k}{\sqrt{2}}-\frac{k-1}{\sqrt{2}} & \sqrt{2} (k-1) \\
   \frac{k-1}{\sqrt{2}}-\frac{k}{\sqrt{2}} & 0 & \frac{k-1}{\sqrt{2}}+\frac{k}{\sqrt{2}} \\
   0 & \frac{k-1}{\sqrt{2}}-\frac{k}{\sqrt{2}} & \frac{k-1}{\sqrt{2}}+\frac{k}{\sqrt{2}} \\
   \end{array}
   \right)$
6. Consider the system of differential equations $\left(\begin{array}{c}x_{1}^{\prime \prime}(t) \\ x_{2}^{\prime \prime}(t) \\ x_{3}^{\prime \prime}(t)\end{array}\right)=\frac{g}{m l}\left(\begin{array}{ccc}-1 & 1 & 0 \\ 1 & -3 & 2 \\ 0 & 2 & -5\end{array}\right)\left(\begin{array}{l}x_{1}(t) \\ x_{2}(t) \\ x_{3}(t)\end{array}\right)$
   which we could abbreviate $\mathbf{x}^{\prime \prime}(t)=\frac{g}{m l} A \mathbf{x}(t)$. These equations provide a simple model of the dynamics of a ‘triple pendulum’, a system of 3 equal masses $m$ hanging under the force of gravity $g$ and connected by strings of length $l$ (see the figure below). The variables $x_{i}(t)$ denote the horizontal displacement.
   Noting that $A$ is symmetric, Spectral Theory gives us that $A$ can be expressed as $A=PDP^{\sf T}$ for an orthogonal matrix $P$ and diagonal matrix$$D=\left(\begin{array}{ccc}-\lambda_{1} & 0 & 0 \\ 0 & -\lambda_{2} & 0 \\ 0 & 0 & -\lambda_{3}\end{array}\right)$$and in this case the $λ_i > 0$. Use this fact to determine the general solution for $\mathbf x(t)$. [You need not explicitly compute $P$ or the $λ_i$.]
   Error! Click to view log.
   Solution.
   Since $A$ is orthogonal and $A=PDP^{\sf T}$, we have $P^{\sf T}\mathbf x''(t)=\frac{g}{ml}DP^{\sf T}\mathbf x(t)$. The solution is $\mathbf x(t) =PC$, where $C=(C_1\;C_2\;C_3)^{\sf T}$, $C_i=c_1 \sin\left(\sqrt{\lambda_ig\over ml} t\right) + c_2 \cos\left(\sqrt{\lambda_ig\over ml} t\right) $