\documentclass{article} %\pagestyle{empty} \setlength{\topmargin}{-.5in} \setlength{\oddsidemargin}{-.3in} \setlength{\textheight}{9in} \setlength{\textwidth}{7.2in} \usepackage{amssymb} \usepackage{amsmath} \newcommand{\adj}{\mathrm{Adj}} \newcommand{\mbf}{\mathbf} \newcommand{\R}{\mathbb{R}} \newcommand{\ra}{\rightarrow} \newcommand{\proj}{\mathrm{proj}} \newcommand{\dotprod}{\bullet} \newcommand{\sol}{\textbf{Solution}} \begin{document} \begin{center} {\Large Key to TEST 2 Math 315} \end{center} \begin{enumerate} \item Consider the matrix $A$ below. \[ A = \left[ \begin{array}{rrrrr} 1 & 0 & 3 & 4 & 0 \\ 0 & 1 & 8 & -1 & 1 \\ 1 & 1 & 11 & 3 & 1 \end{array} \right] \] \begin{enumerate} \item What is the rank of $A$? (justify your answer) \\ \sol: If we perform elimination on $A$ we get: \[ \left[ \begin{array}{rrrrr} 1 & 0 & 3 & 4 & 0 \\ 0 & 1 & 8 & -1 & 1 \\ 1 & 1 & 11 & 3 & 1 \end{array} \right] \stackrel{\mbox{rref}}{\longrightarrow} \left[ \begin{array}{rrrrr} 1 & 0 & 3 & 4 & 0 \\ 0 & 1 & 8 & -1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right] \] And therefore $A$ has rank 2. \item \label{row space} Find a basis for the row space of $A$. (justify your answer) \\ \sol: This means a basis for the row space is $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ \[ \mathbf{v}_1 = \left[ \begin{array}{r} 1 \\ 0 \\ 3 \\ 4 \\ 0 \end{array} \right] \qquad \mathbf{v}_2 = \left[ \begin{array}{r} 0 \\ 1 \\ 8 \\ 1 \\ 1 \end{array} \right] \] \item Find a basis for the column space of $A$. (justify your answer) \\ \sol: from the reduced matrix, we can see that the basis for the column space is the original two columns of the matrix $A$ (and NOT the first 2 columns of the reduced matrix). Therefore, a basis for the column space is: $\{\mathbf{u}_1, \mathbf{u}_2\}$ \[ \mathbf{u}_1 = \left[ \begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right] \qquad \mathbf{u}_2 = \left[ \begin{array}{r} 0 \\ 1 \\ 1 \end{array} \right] \] \item What is the dimension of the null space of $A$? (justify your answer) \\ \sol: Rank nullity theorem tells us that the dimension of the null space is $3$. \item Given that the vector $\mathbf{v}=[8, -5, -16, 37, -5]^t$ is in the row space of $A$, write $\mathbf{v}$ in the coordinates determined by your basis in part~\ref{row space}. \\ \sol: Just look and see that \[ \mathbf{v} = 8\mathbf{v}_1 - 5 \mathbf{v}_2 \] \end{enumerate} \item Consider the subspace of $\R^4$: \[ W = \mathrm{Span}\{ [4,-5,1,1]^t \} \] Let $v=[-12,8,4,-2]^t$. \begin{enumerate} \item Find the orthogonal projection of $\mathbf{v}$ onto the subspace $W$, $\mathrm{Proj}_W(\mathbf{v})$. \\ \sol: let $\mathbf{w}=[4,-5,1,1]^t$ so that $W=\mathrm{Span}\{\mathbf{w}\}$. We just compute using the formula: \[ \mathrm{Proj}_W(\mathbf{v}) = \frac{v\dotprod w}{w\dotprod w} w = \frac{-86}{43}[4,-5,1,1]^t = [-8,10,-2,-2]^t \] \item Determine a vector, $\mathbf{u}$, orthogonal to $W$ so that $\mathrm{Proj}_W(\mathbf{v}) + \mathbf{u} = \mathbf{v}$. Show that the vector you found is orthogonal to $W$. \\ \sol: just subtract and solve for $u$: \[ \mathbf{u} = \mathbf{v} - \mathrm{Proj}_W(\mathbf{v}) = [-4,-2,6,0]^t \] It is an easy check to see that $\mathbf{u} \dotprod \mathbf{w} = 0$. \end{enumerate} \item Let $A$ be an arbitrary $3 \times 2$ matrix. \begin{enumerate} \item $A$ determines a linear transformation from $\R^n$ to $\R^m$, what is $n$ and $m$? \\ \sol: $n=2$ and $m=3$, so the transformation is $\R^2 \ra \R^3$. \item Determine all possibilities for the Rank of $A$ and nullity of $A$. (justify your answer) \\ \sol: I just make a chart of possibilities \begin{tabular}{|c|c|} \hline Rank $A$ & Nullity of $A$ \\ \hline \hline 0 & 2 \\ 1 & 1 \\ 2 & 0 \\ \hline \end{tabular} \item Determine all possibilities for the Rant of $A^t$ and nullity of $A^t$, the transpose of $A$. (justify your answer). \\ \sol: I just make a chart of possibilities \begin{tabular}{|c|c|} \hline Rank $A^t$ & Nullity of $A^t$ \\ \hline \hline 0 & 3 \\ 1 & 2 \\ 2 & 1 \\ \hline \end{tabular} Note that it is not possible for $A^t$ to have rank 3 because there are only 2 rows. \end{enumerate} \item \label{p trans} Suppose $T:\R^2 \ra \R^3$ is a linear transformation such that \[ T\left( \left[ \begin{array}{r} 1 \\ 0 \end{array} \right] \right) = \left[ \begin{array}{r} 1 \\ 2 \\ -5 \end{array} \right] \qquad T\left( \left[ \begin{array}{r} 2 \\ 1 \end{array} \right] \right) = \left[ \begin{array}{r} -1 \\ 0 \\ 1 \end{array} \right] \] \begin{enumerate} \item Write the vector $[0,1]^t$ as a linear combination of $[1,0]^t$ and $[2,1]^t$. \\ (This isn't supposed to be very hard, you should be able to do this without solving equations. But, I certainly won't stop you from solving equations.) \\ \sol: here it is: \[ \left[ \begin{array}{r} 0 \\ 1 \end{array}\right] = -2 \left[ \begin{array}{r} 1 \\ 0 \end{array}\right] + \left[ \begin{array}{r} 2 \\ 1 \end{array}\right] \] \item Using the combination found above, find $T([0,1]^t)$ \\ \sol: we just use that combination and the properties of a linear transformation: \begin{eqnarray*} T\left( \left[ \begin{array}{r} 0 \\ 1 \end{array}\right] \right) & = & T \left( -2 \left[ \begin{array}{r} 1 \\ 0 \end{array}\right] + \left[ \begin{array}{r} 2 \\ 1 \end{array}\right] \right) \\ & = & -2 T \left( \left[ \begin{array}{r} 1 \\ 0 \end{array}\right] \right) + T \left( \left[ \begin{array}{r} 2 \\ 1 \end{array}\right] \right) \\ & = & -2 \left[ \begin{array}{r} 1 \\ 2 \\ -5 \end{array} \right] + \left[ \begin{array}{r} -1 \\ 0 \\ 1 \end{array} \right] \\ & = & \left[ \begin{array}{r} -3 \\ -4 \\ 11 \end{array} \right] \end{eqnarray*} \item \label{trans} Using this information, find a matrix that represents the transformation $T$. \\ \sol: We know that the columns of the matrix of the transformation are given by the transformation applied to the standard basis elements, which we computed above. Therefore, the matrix is \[ \left[ \begin{array}{rr} 1 & -3 \\ 2 & -4 \\ -5 & 11 \end{array} \right] \] \end{enumerate} \item Consider the standard basis for $\R^2$ and the basis $B=\{ [1,0]^t, [2,1]^t \}$. \begin{enumerate} \item \label{p matrix} Find the change of coordinate matrix to transform coordinate in $B$ to standard coordinates. Call this matrix $P$. (the point matrix using the text's language). \\ \sol: this is just the matrix with these vectors as columns: \[ P = \left[ \begin{array}{rr} 1 & 2 \\ 0 & 1 \end{array} \right] \] \item Find the change of coordinate matrix to transform standard coordinates to coordinates in $B$. Call this matrix $C$. (the coordinate matrix using the text's language). \\ \sol: $C=P^{-1}$, so we just find the inverse using elimination: \[ C = \left[ \begin{array}{rr} 1 & -2 \\ 0 & 1 \end{array} \right] \] \end{enumerate} \item (Extra credit - 5 points) Let $A$ be the matrix from part~\ref{trans} and $P$ the matrix from part~\ref{p matrix}. \begin{enumerate} \item Determine the product $AP$. \\ \sol: this is just a matter of computing: \[ AP = \left[ \begin{array}{rr} 1 & -3 \\ 2 & -4 \\ -5 & 11 \end{array} \right] \left[ \begin{array}{rr} 1 & 2 \\ 0 & 1 \end{array} \right] = \left[ \begin{array}{rr} 1 & -1 \\ 2 & 0 \\ -5 & 1 \end{array} \right] \] \item $AP$ should be a matrix with columns equal to the vectors in problem~\ref{p trans}. Explain why this happens. \\ \sol: What is happening is we are computing the matrix of the transformation in a different basis. The easiest way to understand this is with a diagram (that I am having trouble typing into the computer). \end{enumerate} \end{enumerate} \end{document}