Commit 8aef79a2 by Jim Hefferon

### edits for det2

parent baae0888
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 ... ... @@ -22756,10 +22756,10 @@ octave:6> gplot z \begin{ans}{Four.II.1.9} Solving \begin{equation*} c_1\colvec{3 \\ 3 \\ 1} +c_2\colvec{2 \\ 6 \\ 1} +c_3\colvec{1 \\ 0 \\ 5} =\colvec{4 \\ 1 \\ 2} c_1\colvec[r]{3 \\ 3 \\ 1} +c_2\colvec[r]{2 \\ 6 \\ 1} +c_3\colvec[r]{1 \\ 0 \\ 5} =\colvec[r]{4 \\ 1 \\ 2} \end{equation*} gives the unique solution $$c_3=11/57$$, $$c_2=-40/57$$ and $$c_1=99/57$$. ... ... @@ -22771,8 +22771,8 @@ octave:6> gplot z so that it becomes the box formed by \begin{equation*} \sequence{ \colvec{3 \\ 0}, \colvec{2 \\ 1} \colvec[r]{3 \\ 0}, \colvec[r]{2 \\ 1} } \end{equation*} and now the absolute value of this determinant is ... ... @@ -22815,32 +22815,32 @@ octave:6> gplot z \end{ans} \begin{ans}{Four.II.1.15} For a box we take a sequence of vectors (as described in the remark, the order in which the vectors are taken matters), in the remark, the order of the vectors matters), while for a span we take a set of vectors. Also, for a box subset of $\Re^n$ there must be $n$ vectors; of course for a span there can be any number of vectors. Finally, for a box the coefficients $t_1$,~\ldots, $t_n$ are restricted to the interval $[0..1]$, while for a are in the interval $[0..1]$, while for a span the coefficients are free to range over all of $\Re$. \end{ans} \begin{ans}{Four.II.1.16} That picture is drawn to mislead. We have drawn that picture to mislead. The picture on the left is not the box formed by two vectors. If we slide it to the origin then it becomes the box formed by this sequence. \begin{equation*} \sequence{ \colvec{0 \\ 1}, \colvec{2 \\ 0} \colvec[r]{0 \\ 1}, \colvec[r]{2 \\ 0} } \end{equation*} Then the image under the action of the matrix is the box formed by this sequence. \begin{equation*} \sequence{ \colvec{1 \\ 1}, \colvec{4 \\ 0} \colvec[r]{1 \\ 1}, \colvec[r]{4 \\ 0} } \end{equation*} which has an area of $4$. ... ... @@ -22870,10 +22870,10 @@ octave:6> gplot z \begin{ans}{Four.II.1.20} No, for instance the determinant of \begin{equation*} T=\begin{pmatrix} T=\begin{mat}[r] 2 &0 \\ 0 &1/2 \end{pmatrix} \end{mat} \end{equation*} is $$1$$ so it preserves areas, but the vector $$T\vec{e}_1$$ has length $$2$$. ... ... @@ -22886,21 +22886,21 @@ octave:6> gplot z \begin{ans}{Four.II.1.22} Two of the three sides of the triangle are formed by these vectors. \begin{equation*} \colvec{2 \\ 2 \\ 2}-\colvec{1 \\ 2 \\ 1}=\colvec{1 \\ 0 \\ 1} \colvec[r]{2 \\ 2 \\ 2}-\colvec[r]{1 \\ 2 \\ 1}=\colvec[r]{1 \\ 0 \\ 1} \qquad \colvec{3 \\ -1 \\ 4}-\colvec{1 \\ 2 \\ 1}=\colvec{2 \\ -3 \\ 3} \colvec[r]{3 \\ -1 \\ 4}-\colvec[r]{1 \\ 2 \\ 1}=\colvec[r]{2 \\ -3 \\ 3} \end{equation*} One way to find the area of this triangle is to produce a length-one vector orthogonal to these two. From these two relations \begin{equation*} \colvec{1 \\ 0 \\ 1} \colvec[r]{1 \\ 0 \\ 1} \cdot\colvec{x \\ y \\ z} =\colvec{0 \\ 0 \\ 0} =\colvec[r]{0 \\ 0 \\ 0} \qquad \colvec{2 \\ -3 \\ 3} \colvec[r]{2 \\ -3 \\ 3} \cdot\colvec{x \\ y \\ z} =\colvec{0 \\ 0 \\ 0} =\colvec[r]{0 \\ 0 \\ 0} \end{equation*} we get a system \begin{equation*} ... ... @@ -22916,11 +22916,11 @@ octave:6> gplot z \end{equation*} with this solution set. \begin{equation*} \set{\colvec{-1 \\ 1/3 \\ 1}z\suchthat z\in\Re}, \set{\colvec[r]{-1 \\ 1/3 \\ 1}z\suchthat z\in\Re}, \end{equation*} A solution of length one is this. \begin{equation*} \frac{1}{\sqrt{19/9}}\colvec{-1 \\ 1/3 \\ 1} \frac{1}{\sqrt{19/9}}\colvec[r]{-1 \\ 1/3 \\ 1} \end{equation*} Thus the area of the triangle is the absolute value of this determinant. ... ... @@ -22965,7 +22965,7 @@ octave:6> gplot z as the determinant $$\deter{TS}$$ changes sign on the row swap, we'll then have $$\deter{\hat{T}S}=-\deter{TS}$$, and so $$d(\hat{T})=-d(T)$$). This ckeck runs just like the one for the first property. This check runs just like the one for the first property. For the third property, we need only show that performing $T\smash[b]{\grstep{k\rho_i}}\hat{T}$ ... ... @@ -22995,10 +22995,10 @@ octave:6> gplot z The converse does not hold; here is an example. \begin{equation*} \begin{pmatrix} \begin{mat}[r] 3 &1 \\ 2 &1 \end{pmatrix} \end{mat} \end{equation*} \end{ans} ... ... @@ -23019,14 +23019,14 @@ octave:6> gplot z \partsitem The new basis is the old basis rotated by $$\pi/4$$. \partsitem $\sequence{\colvec{-1 \\ 0}, \colvec{0 \\ -1}} \sequence{\colvec[r]{-1 \\ 0}, \colvec[r]{0 \\ -1}}$, $\sequence{\colvec{0 \\ -1}, \colvec{1 \\ 0}} \sequence{\colvec[r]{0 \\ -1}, \colvec[r]{1 \\ 0}}$ \partsitem In each case the determinant is $$+1$$ (these bases are said to (we say that these bases have \definend{positive orientation}). \partsitem Because only one sign can change at a time, the only other cycle possible is ... ... @@ -23039,7 +23039,7 @@ octave:6> gplot z \;\longrightarrow\cdots\,. \end{equation*} Here each associated determinant is $$-1$$ (such bases are said to have a \definend{negative orientation}). (we say that such bases have a \definend{negative orientation}). \partsitem There is one positively oriented basis $$\sequence{(1)}$$ and one negatively oriented basis $$\sequence{(-1)}$$. \partsitem There are $$48$$ bases ($$6$$ half-axis choices are ... ... @@ -23054,8 +23054,9 @@ octave:6> gplot z \includegraphics{ch4.48} \end{center} In $$\Re^3$$ positive orientation is sometimes called right hand orientation' because if a person's right hand is placed with the fingers curling right hand orientation' because if a person places their right hand with their fingers curling from $$\vec{e}_1$$ to $$\vec{e}_2$$ then the thumb will point with $$\vec{e}_3$$. \end{exparts} ... ... @@ -23145,7 +23146,7 @@ octave:6> gplot z \end{equation*} As in the proof that the determinant of a matrix equals the determinant of its transpose, we commute the $s$'s so they are listed by ascending of its transpose, we commute the $s$'s to list them by ascending row number instead of by ascending column number (and we substitute $\sgn(\phi^{-1})$ for $\sgn(\phi)$). \begin{equation*} ... ... @@ -23184,8 +23185,8 @@ octave:6> gplot z That only happens (assuming that $(x_2,y_3)\neq (x_3,y_3)$) if $(x,y)$ lies on the line through the other two. \partsitem \answerasgiven % The altitude through $(x_1,y_1)$ of a triangle with vertices $(x_1,y_1)$ $(x_2,y_2)$ and $(x_3,y_3)$ is found in the usual We find the altitude through $(x_1,y_1)$ of a triangle with vertices $(x_1,y_1)$ $(x_2,y_2)$ and $(x_3,y_3)$ in the usual way from the normal form of the above: \begin{equation*} \frac{1}{\sqrt{(x_2-x_3)^2+(y_2-y_3)^2}} ... ... @@ -23205,7 +23206,7 @@ octave:6> gplot z \end{vmatrix}. \end{equation*} This exposition reveals the \textit{modus operandi} more clearly than the usual proof of showing a collection of terms to be identitical than the usual proof of showing a collection of terms to be identical with the determinant. \partsitem \answerasgiven % Let
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 ... ... @@ -14,7 +14,7 @@ input graph % GLOBAL Values color lightgray; lightgray=.70white; lightgray=.80white; color medgray; medgray=.60white; color darkgray; ... ...
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