\documentclass[12pt]{article} \usepackage{amsfonts} %\documentstyle[12pt,amsfonts]{article} %\documentstyle{article} \setlength{\topmargin}{-.5in} \setlength{\oddsidemargin}{0 in} \setlength{\evensidemargin}{0 in} \setlength{\textwidth}{6.5truein} \setlength{\textheight}{8.5truein} %\input ../basicmath/basicmathmac.tex % %\input ../adgeomcs/lamacb.tex \input ../adgeomcs/mac.tex \input ../adgeomcs/mathmac.tex \def\fseq#1#2{(#1_{#2})_{#2\geq 1}} \def\fsseq#1#2#3{(#1_{#3(#2)})_{#2\geq 1}} \def\qleq{\sqsubseteq} % \begin{document} \begin{center} \fbox{{\Large\bf Spring, 2005 \hspace*{0.4cm} CIS 610}}\\ \vspace{1cm} {\Large\bf Advanced Geometric Methods in Computer Science\\ Jean Gallier \\ \vspace{0.5cm} Homework 2}\$10pt] February 23, 2005; Due March 16, 2005\\ (Note the new due date!) \end{center} A problems'' are for practice only, and should not be turned in. \vspace {0.25cm}\noindent {\bf Problem A1.} (a) Find two symmetric matrices, A and B, such that AB is not symmetric. \medskip (b) Find two matrices, A and B, such that \[ e^Ae^B \not= e^{A + B}.$ Try $A = \frac{\pi}{2}\mattc{0}{0}{0} {0}{0}{-1} {0}{1}{0} \quad\hbox{and}\quad B = \frac{\pi}{2}\mattc{0}{0}{1} {0}{0}{0} {-1}{0}{0}.$ \vspace {0.25cm}\noindent {\bf Problem A2.} (a) If $K = \reals$ or $K = \complex$, recall that the projective space, $\projs{K^{n+1}}$, is the set of equivalence classes of the equivalence relation, $\sim$, on $K^{n + 1} - \{0\}$, defined so that, for all $u, v \in K^{n + 1} - \{0\}$, $u \sim v\quad\hbox{iff}\quad v = \lambda u, \quad\hbox{for some}\quad\lambda \in K - \{0\}.$ The map, $\mapdef{p}{(K^{n+1} - \{0\})}{\projs{K^{n+1}}}$, is the projection mapping any nonzero vector in $K^{n+1}$ to its equivalence class modulo $\sim$. We let $\rprospac{n} = \projs{\reals^{n+1}}$ and $\cprospac{n} = \projs{\complex^{n+1}}$. \medskip Prove that for any $n\geq 0$, there is a bijection between $\projs{K^{n+1}}$ and $K^{n} \cup \projs{K^{n}}$ (which allows us to identify them). \medskip (b) Prove that $\rprospac{n}$ and $\cprospac{n}$ are connected and compact. \medskip \hint If $S^n = \{(x_1,\ldots,x_{n+1})\in K^{n+1}\ |\ x_1^2 + \cdots + x_{n+1}^2 = 1\},$ prove that $p(S^n) = \projs{K^{n+1}}$, and recall that $S^n$ is compact for all $n\geq 0$ and connected for $n\geq 1$. For $n = 0$, $\projs{K}$ consists of a single point. \vspace {0.25cm}\noindent {\bf Problem A3.} Recall that $\reals^2$ and $\complex$ can be identified using the bijection $(x, y) \mapsto x + iy$. Also recall that the subset $U(1)\subseteq \complex$ consisting of all complex numbers of the form $\cos\theta + i\sin\theta$ is homeomorphic to the circle $S^1 = \{(x, y)\in \reals^2\ |\ x^2 + y^2 = 1\}$. If $\mapdef{c}{U(1)}{U(1)}$ is the map defined such that $c(z) = z^2,$ prove that $c(z_1) = c(z_2)$ iff either $z_2 = z_1$ or $z_2 = -z_1$, and thus that $c$ induces a bijective map $\mapdef{\hli{c}}{\rprospac{1}}{S^1}$. Prove that $\hli{c}$ is a homeomorphism (remember that $\rprospac{1}$ is compact). \vspace {0.5cm} B problems'' must be turned in. \vspace {0.25cm}\noindent {\bf Problem B1 (20 pts).} Let $A = (a_{i\, j})$ be a real or complex $n\times n$ matrix. \medskip (1) If $\lambda$ is an eigenvalue of $A$, prove that there is some eigenvector $\novect{u} = (u_1, \ldots, u_n)$ of $A$ for $\lambda$ such that $\max_{1\leq i\leq n} |u_i| = 1.$ \medskip (2) If $\novect{u} = (u_1, \ldots, u_n)$ is an eigenvector of $A$ for $\lambda$ as in (1), assuming that $i$, $1\leq i \leq n$, is an index such that $|u_i| = 1$, prove that $(\lambda - a_{i\, i})u_i = \sum_{{j = 1} \atop {j\not= i}}^{n} a_{i\, j} u_j,$ and thus that $|\lambda - a_{i\, i}| \leq \sum_{{j = 1} \atop {j\not= i}}^{n} |a_{i\, j}|.$ Conclude that the eigenvalues of $A$ are inside the union of the closed disks $D_i$ defined such that $D_i = \biggl\{z\in \complex\ |\ |z - a_{i\, i}| \leq \sum_{{j = 1} \atop {j\not= i}}^{n} |a_{i\, j}|\biggr\}.$ \medskip \remark This result is known as {\it Gershgorin's theorem\/}. \vspace {0.5cm}\noindent \vspace {0.25cm}\noindent {\bf Problem B2 (10).} Recall that a real $n\times n$ symmetric matrix, $A$, is {\it positive semi-definite\/} iff its eigenvalues, $\lambda_1, \ldots, \lambda_n$ are non-negative (i.e., $\lambda_i \geq 0$ for $i = 1, \ldots, n$) and {\it positive definite\/} iff its eigenvalues are positive (i.e., $\lambda_i >0$ for $i = 1, \ldots, n$). \medskip (a) Prove that a symmetric matrix, $A$, is positive semi-definite iff $\transpos{X}A X \geq 0$, for all $X\not= 0$ ($X\in \reals^n$) and positive definite iff $\transpos{X}A X > 0$, for all $X\not= 0$ ($X\in \reals^n$). \medskip (b) Prove that for any two positive definite matrices, $A, B$, for all $\lambda, \mu \in \reals$, with $\lambda, \mu \geq 0$ and $\lambda + \mu > 0$, the matrix $\lambda A + \mu B$ is still symmetric, positive definite. Deduce that the set of $n\times n$ symmetric positive definite matrices is convex (in fact, a cone). \vspace {0.25cm}\noindent {\bf Problem B3 (40).} \label{pb5.3} (i) In $\reals^3$, the sphere $S^2$ is the set of points of coordinates $(x, y, z)$ such that $x^2 + y^ 2 + z^2 = 1$. The point $N = (0, 0, 1)$ is called the {\it north pole\/}, and the point $S = (0, 0, -1)$ is called the {\it south pole\/}. The {\it stereographic projection map\/} $\mapdef{\sigma_N}{(S^2 - \{N\})}{\reals^2}$ is defined as follows: For every point $M\not= N$ on $S^2$, the point $\sigma_N(M)$ is the intersection of the line through $N$ and $M$ and the plane of equation $z = 0$. Show that if $M$ has coordinates $(x, y, z)$ (with $x^2 + y^2 + z^2 = 1$), then $\sigma_N(M) = \biggl(\frac{x}{1 - z},\, \frac{y}{1 - z}\biggr).$ Prove that $\sigma_N$ is bijective and that its inverse is given by the map $\mapdef{\tau_N}{\reals^2}{(S^2 - \{N\})}$, with $(x, y) \mapsto \biggl(\frac{2x}{x^2 + y^2 + 1},\, \frac{2y}{x^2 + y^2 + 1},\, \frac{x^2 + y^2 - 1}{x^2 + y^2 + 1}\biggr).$ Similarly, $\mapdef{\sigma_S}{(S^2 - \{S\})}{\reals^2}$ is defined as follows: For every point $M\not= S$ on $S^2$, the point $\sigma_S(M)$ is the intersection of the line through $S$ and $M$ and the plane of equation $z = 0$. Show that $\sigma_S(M) = \biggl(\frac{x}{1 + z},\, \frac{y}{1 + z}\biggr).$ Prove that $\sigma_S$ is bijective and that its inverse is given by the map $\mapdef{\tau_S}{\reals^2}{(S^2 - \{S\})}$, with $(x, y) \mapsto \biggl(\frac{2x}{x^2 + y^2 + 1},\, \frac{2y}{x^2 + y^2 + 1},\, \frac{1 - x^2 - y^2}{x^2 + y^2 + 1}\biggr).$ \medskip Using the complex number $u = x + i y$ to represent the point $(x, y)$, the maps $\mapdef{\tau_N}{\reals^2}{(S^2 - \{N\})}$ and $\mapdef{\sigma_N}{(S^2 - \{N\})}{\reals^2}$ can be viewed as maps from $\complex$ to $(S^2 - \{N\})$ and from $(S^2 - \{N\})$ to $\complex$, defined such that $\tau_N(u) = \biggl(\frac{2u}{|u|^2 + 1},\, \frac{|u|^2 - 1}{|u|^2 + 1}\biggr)$ and $\sigma_N(u, z) = \frac{u}{1 - z},$ and similarly for $\tau_S$ and $\sigma_S$. Prove that if we pick two suitable orientations for the $xy$-plane, we have $\sigma_N(M)\sigma_S(M) = 1,$ for every $M \in S^2 - \{N, S\}$. \medskip (ii) Identifying $\complex^2$ and $\reals^4$, for $z = x + i y$ and $z' = x' + i y'$, we define $\smnorme{(z, z')} = \sqrt{x^2 + y^2 + x'^2 + y'^2}.$ The sphere $S^3$ is the subset of $\complex^2$ (or $\reals^4$) consisting of those points $(z, z')$ such that $\smnorme{(z, z')}^2 = 1$. \medskip Prove that $\projs{\complex^2} = p(S^3)$, where $\mapdef{p}{(\complex^2 - \{(0, 0)\})}{\projs{\complex^2}}$ is the projection map. If we let $u = z/z'$ (where $z, z'\in \complex$) in the map $u \mapsto \biggl(\frac{2u}{|u|^2 + 1},\, \frac{|u|^2 - 1}{|u|^2 + 1}\biggr)$ and require that $\smnorme{(z, z')}^2 = 1$, show that we get the map $\mapdef{HF}{S^3}{S^2}$ defined such that $HF((z, z')) = (2 z\overline{z'},\, |z|^2 - |z'|^2).$ Prove that $\mapdef{HF}{S^3}{S^2}$ induces a bijection $\mapdef{\hli{HF}}{\projs{\complex^2}}{S^2}$, and thus that $\cprospac{1} = \projs{\complex^2}$ is homeomorphic to $S^2$. \medskip (iii) Prove that the inverse image $HF^{-1}(s)$ of every point $s\in S^2$ is a circle. Thus $S^3$ can be viewed as a union of disjoint circles. The map $HF$ is called the {\it Hopf fibration\/}. \vspace {0.25cm}\noindent {\bf Problem B4 (60).} (a) Consider the map $\mapdef{\s{H}}{\reals^3}{\reals^4}$ defined such that $(x, y, z) \mapsto (xy, yz, xz, x^2 - y^2).$ Prove that when it is restricted to the sphere $S^2$ (in $\reals^3$), we have $\s{H}(x, y, z) = \s{H}(x', y', z')$ iff $(x', y', z') = (x, y, z)$ or $(x', y', z') = (-x, -y, -z)$. In other words, the inverse image of every point in $\s{H}(S^2)$ consists of two antipodal points. \medskip Prove that the map $\s{H}$ induces an injective map from the projective plane onto $\s{H}(S^2)$, and that it is a homeomorphism. \medskip (b) The map $\s{H}$ allows us to realize concretely the projective plane in $\reals^4$ as an embedded manifold. Consider the three maps from $\reals^2$ to $\reals^4$ given by \begin{eqnarray*} \psi_1(u, v) & = & \left( \frac{uv}{u^2 + v^2 + 1}, \frac{u}{u^2 + v^2 + 1}, \frac{v}{u^2 + v^2 + 1}, \frac{u^2 - v^2}{u^2 + v^2 + 1} \right), \\ \psi_2(u, v) & = & \left( \frac{u}{u^2 + v^2 + 1}, \frac{v}{u^2 + v^2 + 1}, \frac{uv}{u^2 + v^2 + 1}, \frac{u^2 - 1}{u^2 + v^2 + 1} \right), \\ \psi_3(u, v) & = & \left( \frac{u}{u^2 + v^2 + 1}, \frac{uv}{u^2 + v^2 + 1}, \frac{v}{u^2 + v^2 + 1}, \frac{1 - u^2}{u^2 + v^2 + 1} \right). \end{eqnarray*} % Observe that $\psi_1$ is the composition $\s{H} \circ \alpha_1$, where $\alpha_1\co \reals^2 \longrightarrow S^2$ is given by $(u, v) \mapsto \left(\frac{u}{\sqrt{u^2 + v^2 + 1}}, \frac{v}{\sqrt{u^2 + v^2 + 1}}, \frac{1}{\sqrt{u^2 + v^2 + 1}} \right),$ that $\psi_2$ is the composition $\s{H} \circ \alpha_2$, where $\alpha_2\co \reals^2 \longrightarrow S^2$ is given by $(u, v) \mapsto \left(\frac{u}{\sqrt{u^2 + v^2 + 1}}, \frac{1}{\sqrt{u^2 + v^2 + 1}}, \frac{v}{\sqrt{u^2 + v^2 + 1}} \right).$ and $\psi_3$ is the composition $\s{H} \circ \alpha_3$, where $\alpha_3\co \reals^2 \longrightarrow S^2$ is given by $(u, v) \mapsto \left(\frac{1}{\sqrt{u^2 + v^2 + 1}}, \frac{u}{\sqrt{u^2 + v^2 + 1}}, \frac{v}{\sqrt{u^2 + v^2 + 1}} \right),$ Prove that each $\psi_i$ is injective, continuous and nonsingular (i.e., the Jacobian is never zero). \medskip Prove that $\psi_i(\reals^2)$ is an open subset, $U_i$, of $\s{H}(S^2)$ for $i = 1, 2, 3$ and that the union of the $U_i$'s covers $\s{H}(S^2)$. \medskip Prove that each $\mapdef{\psi_i^{-1}}{U_i}{\reals^2}$ is continuous. This is a little tricky. For example, for $\psi_1$, first prove that if the coordinates in $\reals^4$ are $(x, y, z, t)$, then $yzt = x(y^2 - z^2).$ Then, $\psi_1^{-1}$ is defined as follows: If $y\not= 0$ and $z \not= 0$, $u = \frac{x}{z} = \frac{yt}{y^2 - z^2}, \quad v = \frac{x}{y} = \frac{zt}{y^2 - z^2}.$ If $y = 0$ and $z \not= 0$, then $u = 0, \quad v = -\frac{t}{z},$ if $y\not= 0$ and $z = 0$, then $u = \frac{t}{y}\quad v = 0,$ and if $y = z = 0$, then $u = 0,\quad v = 0.$ Finally, you have to show continuity of the above functions, and do a similar thing for $\psi_2^{-1}$ and $\psi_3^{-1}$. \medskip Conclude that $\psi_1, \psi_2, \psi_3$ are parametrizations of $\rprospac{2}$ as a manifold in $\reals^4$. \medskip (c) Investigate the surfaces in $\reals^3$ obtained by dropping one of the four coordinates. Show that there are only two of them (the Steiner Roman surface'' and the crosscap'', up to a rigid motion). \vspace {0.25cm}\noindent {\bf Problem B5 (20).} (a) Let $A$ be any invertible (real) $n\times n$ matrix. Prove that for every SVD, $A = VD\transpos{U}$, of $A$, the product $V\transpos{U}$ is the same (i.e., if $V_1D\transpos{U_1} = V_2D\transpos{U_2}$, then $V_1\transpos{U_1} = V_2\transpos{U_2}$). What does $V\transpos{U}$ have to do with the polar form of $A$? \medskip (b) Given any invertible (real) $n\times n$ matrix, $A$, prove that there is a unique orthogonal matrix, $Q\in \mathbf{O}(n)$, such that $\norme{A - Q}_F$ is minimal (under the Frobenius norm). In fact, prove that $Q = V\transpos{U}$, where $A = VD\transpos{U}$ is an SVD of $A$. Moreover, if $\det(A) > 0$, show that $Q\in \mathbf{SO}(n)$. \medskip What can you say if $A$ is singular (i.e., non-invertible)? \vspace {0.25cm}\noindent {\bf Problem B6 (40).} (a) Consider the map, $\mapdef{f}{\mathbf{GL}^+(n)}{\mathbf{S}(n)}$, given by $f(A) = \transpos{A}{A} - I.$ Check that $df(A)(H) = \transpos{A}H + \transpos{H}A,$ for any matrix, $H$. \medskip (b) Consider the map, $\mapdef{f}{\mathbf{GL}(n)}{\reals}$, given by $f(A) = \det(A).$ Prove that $df(I)(B) = \mathrm{tr}(B)$, the trace of $B$, for any matrix $B$ (here, $I$ is the identity matrix). Then, prove that $df(A)(B) = \det(A)\mathrm{tr}(A^{-1}B),$ where $A\in \mathbf{GL}(n)$. \medskip (c) Use the map $A \mapsto \det(A) - 1$ to prove that $\mathbf{SL}(n)$ is a manifold of dimension $n^2 - 1$. \medskip (d) Let $J$ be the $(n+1)\times (n+1)$ diagonal matrix $J = \matta{I_n}{0}{0}{-1}.$ We denote by $\mathbf{SO}(n, 1)$ the group of real $(n+1)\times (n+1)$ matrices $\mathbf{SO}(n, 1) = \{A\in \mathbf{GL}(n+1) \mid \transpos{A}JA = J\quad\hbox{and}\quad \det(A) = 1\}.$ Check that $\mathbf{SO}(n, 1)$ is indeed a group with the inverse of $A$ given by $A^{-1} = J\transpos{A}J$ (this is the {\it special Lorentz group\/}.) Consider the function $\mapdef{f}{\mathbf{GL}^{+}(n+1)}{\mathbf{S}(n+1)}$, given by $f(A) = \transpos{A}JA - J,$ where $\mathbf{S}(n+1)$ denotes the space of $(n+1)\times (n+1)$ symmetric matrices. Prove that $df(A)(H) = \transpos{A}JH + \transpos{H}JA$ for any matrix, $H$. Prove that $df(A)$ is surjective for all $A\in \mathbf{SO}(n, 1)$ and that $\mathbf{SO}(n, 1)$ is a manifold of dimension $\frac{n(n+1)}{2}$. \vspace {0.25cm}\noindent {\bf Problem B7 (40 pts).} (a) Given any matrix $B = \matta{a}{b}{c}{-a}\in \mfrac{sl}(2, \complex),$ if $\omega^2 = a^2 + bc$ and $\omega$ is any of the two complex roots of $a^2 + bc$, prove that if $\omega\not= 0$, then $e^B = \cosh \omega\, I + \frac{\sinh\, \omega}{\omega}\, B,$ and $e^B = I + B$, if $a^2 + bc = 0$. Observe that $\mathrm{tr}(e^B) = 2\cosh\,\omega$. \medskip Prove that the exponential map, $\mapdef{\exp}{\mfrac{sl}(2, \complex)}{\mathbf{SL}(2, \complex)}$, is {\it not\/} surjective. For instance, prove that $\matta{-1}{1}{0}{-1}$ is not the exponential of any matrix in $\mfrac{sl}(2, \complex)$. \medskip (b) Recall that a matrix, $N$, is {\it nilpotent\/} iff there is some $m\geq 0$ so that $N^m = 0$. Let $A$ be any $n\times n$ matrix of the form $A = I - N$, where $N$ is nilpotent. Why is $A$ invertible? prove that there is some $B$ so that $e^B = I - N$ as follows: Recall that for any $y\in\reals$ so that $|y - 1|$ is small enough, we have $\log(y) = -(1 - y) - \frac{(1 - y)^2}{2} - \cdots - \frac{(1 - y)^k}{k} - \cdots.$ As $N$ is nilpotent, we have $N^m = 0$, where $m$ is the smallest integer with this propery. Then, the expression $B = \log(I - N) = -N - \frac{N^2}{2} - \cdots - \frac{N^{m-1}}{m-1}$ is well defined. Use a formal power series argument to show that $e^{B} = A.$ We denote $B$ by $\log(A)$. \medskip (c) Let $A\in \mathbf{GL}(n, \complex)$. Prove that there is some matrix, $B$, so that $e^B = A$. Thus, the exponential map, $\mapdef{\exp}{\mfrac{gl}(n, \complex)}{\mathbf{GL}(n, \complex)}$, is surjective. \medskip First, use the fact that $A$ has a Jordan form, $PJP^{-1}$. Then, show that finding a log of $A$ reduces to finding a log of every Jordan block of $J$. As every Jordan block, $J$, has a fixed nonzero constant, $\lambda$, on the diagonal, with $1$'s immediately above each diagonal entry and zero's everywhere else, we can write $J$ as $(\lambda I)(I - N)$, where $N$ is niplotent. Find $B_1$ and $B_2$ so that $\lambda I = e^{B_1}$, $I - N = e^{B_2}$, and $B_1B_2 = B_2B_1$. Conclude that $J = e^{B_1 + B_2}$. \vspace {0.25cm}\noindent {\bf Problem B8 (50 pts).} Recall that for any matrix $A = \mattc{0}{-c}{b}{c}{0}{-a}{-b}{a}{0},$ if we let $\theta = \sqrt{a^2 + b^2 + c^2}$ and % $B = \mattc{a^2}{ab}{ac}{ab}{b^2}{bc}{ac}{bc}{c^2},$ then the exponential map, $\mapdef{\exp}{{\mfrac{so}}(3)}{\mSO{3}}$, is given by $\exp{A} = e^A = \cos\theta\, I_3 + \frac{\sin\theta}{\theta} A + \frac{(1 - \cos\theta)}{\theta^2} B,$ or, equivalently, by $e^A = I_3 + \frac{\sin\theta}{\theta} A + \frac{(1 - \cos\theta)}{\theta^2} A^2,$ if $\theta \not= k2\pi$ ($k\in\integs$), with $\exp(0_3) = I_3$ (Rodrigues's formula (1840)) \medskip (a) Let $R\in \mathbf{SO}(3)$ and assume that $R\not= I$ and $\mathrm{tr}(R) \not= -1$. Then, prove that a log of $R$ (i.e., a skew symmetric matrix, $S$, so that $e^S = R$) is given by $\log(R) = \frac{\theta}{2\sin\theta}(R - R^T),$ where $1 + 2\cos\theta = \trace(R)$ and $0 < \theta < \pi$. \medskip (b) Now, assume that $\mathrm{tr}(R) = -1$. In this case, show that $R$ is a rotation of angle $\pi$, that $R$ is symmetric and has eigenvalues, $-1, -1, 1$. Assuming that $e^A = R$, Rodrigues formula becomes $R = I + \frac{2}{\pi^2} A^2,$ so $A^2 = \frac{\pi^2}{2}(R - I).$ If we let $S = A/\pi$, we see that we need to find a skew-symmetric matrix, $S$, so that $S^2 = \frac{1}{2}(R - I) = C.$ Observe that $C$ is also symmetric and has eigenvalues, $-1, -1, 0$. Thus, we can diagonalize $C$, as $C = P\mattc{-1}{0}{0} {0}{-1}{0} {0}{0}{0} \transpos{P},$ and if we let $S = P\mattc{0}{-1}{0} {1}{0}{0} {0}{0}{0} \transpos{P},$ check that $S^2 = C$. \medskip (c) From (a) and (b), we know that we can compute explicity a log of a rotation matrix, although when $\theta \approx 0$, we have to be careful in computing $\frac{\sin\theta}{\theta}$; in this case, we may want to use $\frac{\sin\theta}{\theta} = 1 - \frac{\theta^2}{3!} + \frac{\theta^4}{5!} + \cdots.$ Given two rotations, $R_1, R_2\in \mathbf{SO}(3)$, there are three natural interpolation formulae: $e^{(1 - t)\log R_1 + t\log R_2}; \quad R_1 e^{t \log(\transpos{R_1}R_2)}; \quad e^{t\log( R_2\transpos{R_1})} R_1,$ with $0 \leq t \leq 1$. \medskip Write a computer program to investigate the difference between these interpolation formulae. The position of a rigid body spinning around its center of gravity is determined by a rotation matrix, $R\in \mathbf{SO}(3)$. If $R_1$ denotes the initial position and $R_2$ the final position of this rigid body, by computing interpolants of $R_1$ and $R_2$, we get a motion of the rigid body and we can create an animation of this motion by displaying several interpolants. The rigid body can be a funny'' object, for example a banana, a bottle, etc. \vspace{0.5cm}\noindent {\bf TOTAL: 280 points.} \end{document}