\documentclass[11pt]{article} \textwidth = 17cm \textheight = 25.5cm \hoffset = -2.8cm \voffset = -2cm \pagestyle{empty} \usepackage{latexsym,amssymb,amsmath} \def\C{\mathbb{C}} \def\R{\mathbb{R}} \def\Z{\mathbb{Z}} \def\Q{\mathbb{Q}} \def\H{\mathbb{H}} \def\N{\mathbb{N}} \def\Ucal{\mathcal{U}} \def\n{\noindent} \begin{document} \begin{center} \textsc{Algebra (MA 183) 2008-2009} \\ \smallskip \textbf{Solutions to Homework Problems 2} \\ \smallskip \end{center} \medskip \begin{enumerate} \item[2.] Let $A=\left(\begin{array}{rr} a_1 & b_1 \\ c_1 & d_1\end{array}\right)$, $B=\left(\begin{array}{rr} a_2 & b_2 \\ c_2 & d_2\end{array}\right)$. Then $$ AB = \left(\begin{array}{rr} a_1a_2+b_1c_2 & a_1b_2+b_1d_2 \\ c_1a_2+d_1c_2 & c_1b_2+d_1d_2\end{array}\right), $$ and \begin{eqnarray*} \det (AB) & = & (a_1a_2+b_1c_2)(c_1b_2+d_1d_2)-(a_1b_2+b_1d_2)(c_1a_2+d_1c_2) \\ & = & a_1a_2d_1d_2 + b_1c_2c_1b_2 - a_1b_2d_1c_2 - b_1d_2c_1a_2 \\ & = & a_1d_1(a_2d_2-b_2c_2)-b_1c_1(a_2d_2-b_2c_2) \\ & = & (a_1d_1-b_1c_1)(a_2d_2-b_2c_2) \\ & = & \det (A) \det (B). \end{eqnarray*} \item[(b)] Suppose that $A$ and $B$ both have inverses, denoted by $A^{-1}B^{_1}$ respectively. Then $$ (AB)(B^{-1}A^{-1}) = A(bb^{-1})A^{-1} = AI_2A^{-1}=AA^{-1}=I_2. $$ Similarly $(B^{-1}A^{-1})(AB)=I_2$, and so $B^{-1}A^{-1}$ is the inverse of the matrix $AB$. \item[3.] \begin{itemize} \item[(a)] A typical point on the line $L$ has coordinates $(t,\frac{4-t}{2})$, where $t$ is a real number. To find the typical coordinates of the image under $T$ of a point of $L$, calculate the matrix product $$ \left(\begin{array}{rr}1 & 4 \\ -2 & 2 \end{array}\right) \left(\begin{array}{r} t \\ \frac{4-t}{2}\end{array}\right) = \left(\begin{array}{r} t+2(4-t) \\ -2t+4-t\end{array}\right) = \left(\begin{array}{r} 8-t \\ 4-3t \end{array}\right) $$ Thus if $(x,y)$ is a point on the image of $L$ under $T$, then for some real number $t$ we have $x=8-t$ and $y=2-t$. Thus $t=8-x$ and $y=4-3(8-x)=3x-20$. So the image of $L$ under $T$ is the line with equation $3x-y=20$. \item[(b)] The equation of the line to which $L$ is mapped by $T$ is the image of $L$ under the inverse of $T$. The matrix of the inverse of $T$ is the inverse of the matrix of $T$. This is $$ \frac{1}{10}\left(\begin{array}{rr}2 & -4 \\ 2 & 1\end{array}\right). $$ We can proceed as above to find the coordinates of a typical point of the image of $L$ under $T^{-1}$. $$ \frac{1}{10}\left(\begin{array}{rr}2 & -4 \\ 2 & 1\end{array}\right)\left(\begin{array}{r} t \\ \frac{4-t}{2}\end{array}\right) = \frac{1}{10}\left(\begin{array}{r}2t-2(4-t) \\ 2t+\frac{4-t}{2}\end{array}\right) = \frac{1}{10}\left(\begin{array}{r} 4t-8 \\ 2+\frac{3}{2}t\end{array}\right). $$ Thus if $(x,y)$ is a point on the image of $L$ under $T^{-1}$, then for some real number $t$ we have $x=\frac{4t-8}{10}$ and $y=\frac{1}{10}(2+\frac{3}{2}t)$. Thus $t=\frac{10x+8}{4}$ and $$ y=\frac{1}{10}\left(2+\frac{3}{2}\left(\frac{10x+8}{4}\right)\right). $$ Simplifying gives $3x-8y+4=0$ : this is the equation of the line which is mapped by $T$ to $L$. \end{itemize} \item[6.] Suppose that $T:\R^n\longrightarrow \R^p$ is a linear transformation with matrix $M_T$. Suppose also that $S:\R^q\longrightarrow\R^m$ is a linear transformation with matrix $M_S$. \begin{enumerate} \item If $p=4$ and $n=6$, how many rows and how many columns are in the matrix $M_T$? \item If $M_S$ is a $5\times 7$ matrix, what is $m$ and what is $q$? \item Under what conditions on $n,p,q,m$ is the composition $T\circ S$ defined? \item Under what conditions on $n,p,q,m$ is the matrix product $M_SM_T$ defined? \item Suppose that $A$ and $B$ are matrices. Explain in your own words why the matrix product $AB$ can be defined only if the number of columns in $A$ is equal to the number of rows in $B$. \end{enumerate} \n\emph{Solution}: \begin{enumerate} \item $4$ rows, $6$ columns. \item $m=5,\ q=7$. \item $T\circ S$ is defined if $m=n$. \item $M_SM_T$ is defined if the composition $S\circ T$ is defined, i.e. if $p=q$. \end{enumerate} \item[7.] \begin{itemize} \item[(a)] The characteristic polynomial of $B$ is $$ \det(\lambda I_2 -B) = \det \left(\begin{array}{cc} \lambda -2 & 3 \\ 1 & \lambda-4\end{array}\right) = (\lambda -2)(\lambda -4)-(3)(1) = \lambda^2-6\lambda +5. $$ \item[(b)] The eigenvalues are the solutions of the equation $\lambda^2-6\lambda +5 = 0$, i.e. $(\lambda -5)(\lambda -1)=0$, eigenvalues $\lambda_1 = 5$, $\lambda_2 = 1$. \item[(c)] For $\lambda_1 = 5$ : $$ \left(\begin{array}{rr} 2 & -3 \\ -1 & 4 \end{array}\right) \left(\begin{array}{r} x \\ y\end{array}\right) = 5 \left(\begin{array}{r} x \\ y\end{array}\right) \Longrightarrow \begin{array}{rcrcr} 2x & - & 3y & = & 5x \\ -x & + & 4y & = & 5y \end{array} $$ Both these equations say $x = -y$; any non-zero vector $\binom{x}{y}$ whose components satisfy this equation is an eigenvector of $B$ corresponding to the eigenvalue 5 : for example $\binom{-1}{1}$. \medskip For $\lambda_2 = 1$ : $$ \left(\begin{array}{rr} 2 & -3 \\ -1 & 4 \end{array}\right) \left(\begin{array}{r} x \\ y\end{array}\right) = 1 \left(\begin{array}{r} x \\ y\end{array}\right) \Longrightarrow \begin{array}{rcrcr} 2x & - & 3y & = & x \\ -x & + & 4y & = & y \end{array} $$ Both these equations say $x = 3y$; any non-zero vector $\binom{x}{y}$ whose components satisfy this equation is an eigenvector of $B$ corresponding to the eigenvalue 1 : for example $\binom{3}{1}$. \item[(d)] $E = \left(\begin{array}{rr} -1 & 3 \\ 1 & 1\end{array}\right),\ \ D = \left(\begin{array}{rr} 5 & 0 \\ 0 & 1\end{array}\right)$. \item[(e)] $B^6 = (EDE^{-1})^6 = ED^6E^{-1}$. Thus \begin{eqnarray*} B^6 & = & \left(\begin{array}{rr} -1 & 3 \\ 1 & 1\end{array}\right) \left(\begin{array}{cc} 5^6 & 0 \\ 0 & 1^6\end{array}\right) \frac{1}{-4}\left(\begin{array}{rr} 1 & -3 \\ -1 & -1\end{array}\right) \\ & = & -\frac{1}{4} \left(\begin{array}{rr} -5^6 & 3 \\ 5^6 & 1\end{array}\right) \left(\begin{array}{rr} 1 & -3 \\ -1 & -1\end{array}\right) \\ & = & -\frac{1}{4} \left(\begin{array}{rr} -5^6-3 & 3(5^6)-3 \\ 5^6-1 & -3(5^6)-1\end{array}\right) \\ & = & \left(\begin{array}{rr} 3907 & -11718 \\ -3906 & 11719 \end{array}\right) \\ \end{eqnarray*} \end{itemize} \item[9.] \begin{enumerate} \item[(a)] This is a matter of straightforward checking. \item[(b)] The recurrence relations can be written in matrix form as $$ \left(\begin{array}{l}F_{n+2} \\F_{n+1}\end{array}\right) = \left(\begin{array}{rr}1 & 1 \\ 0 & 1 \end{array}\right) \left(\begin{array}{l}F_{n+1} \\ F_n\end{array}\right) $$ for $n\ge 0$. \\ Write $A = \left(\begin{array}{rr} 1 & 1 \\ 0 & 1\end{array}\right)$. Then $$ \left(\begin{array}{l}F_{n+1} \\F_{n}\end{array}\right) = A^n \left(\begin{array}{l}F_1 \\ F_0 \end{array}\right) = A^n \left(\begin{array}{l}0 \\ 1 \end{array}\right) $$ So we need to calculate $A^n$. The characteristic polynomial of $A$ is $\lambda^2-\lambda-1$ which has roots $$ a=\frac{1+\sqrt{5}}{2},\ b=\frac{1-\sqrt{5}}{2}. $$ To find an eigenvector of $A$ corresponding to the eigenvalue $a$, we need $$ \left(\begin{array}{rr} 1 & 1 \\ 0 & 1\end{array}\right) \left(\begin{array}{l} x \\ y\end{array}\right) = a \left(\begin{array}{l} x \\ y\end{array}\right) \Longrightarrow \begin{array}{rcrcr} x & + & y & = & ax \\ x & & & = & ay \end{array} $$ Since $a^2-a-1=0$, $\frac{1}{a}=a-1$ and both equations say $a=ay$. Thus $\binom{a}{1}$ is an eigenvector of $A$ corresponding to the eigenvalue $a$. Similarly $\binom{b}{1}$ is an eigenvector of $A$ corresponding to $b$ as an eigenvalue. \\ Thus $A^n=ED^nE^{-1}$ where $$ E=\left(\begin{array}{rr} a & b \\ 1 & 1\end{array}\right),\ D=\left(\begin{array}{rr} a & 0 \\ 0 & b\end{array}\right). $$ Then \begin{eqnarray*} A^n & = & \left(\begin{array}{rr} a & b \\ 1 & 1\end{array}\right)\left(\begin{array}{rr} a^n & 0 \\ 0 & b^n\end{array}\right) \frac{1}{a-b} \left(\begin{array}{rr} 1 & -b \\ -1 & a\end{array}\right) \\ & = & \ \\ & = & \frac{1}{a-b} \left(\begin{array}{rr} a^{n+1} & b^{n+1} \\ a^n & b^n\end{array}\right) \left(\begin{array}{rr} 1 & -b \\ -1 & a\end{array}\right) \\ \ \\ & = & \frac{1}{a-b} \left(\begin{array}{rr} a^{n+1}-b^{n+1} & -ba^{n+1}+ab^{n+1} \\ a^n-b^n & -ba^n+ab^n\end{array}\right) \end{eqnarray*} Now $$ \left(\begin{array}{l}F_{n+1} \\F_{n}\end{array}\right) = A^n\left(\begin{array}{l}1 \\ 0\end{array}\right) = \frac{1}{a-b} \left(\begin{array}{rr} a^{n+1}-b^{n+1} & -ba^{n+1}+ab^{n+1} \\ a^n-b^n & -ba^n+ab^n\end{array}\right) \left(\begin{array}{l}1 \\ 0\end{array}\right) = \left(\begin{array}{l} a^{n+1}-b^{n+1} \\a^n-b^n\end{array}\right). $$ Looking at the second entries here gives $$ F_n = \frac{1}{a-b}(a^n-b^n) = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right), $$ as required. \end{enumerate} \end{enumerate} \end{document}