full edits for map6

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 ... ... @@ -17249,8 +17249,8 @@ =\vec{v} \end{equation*} (\textit{Remark}. If we assume that $\vec{v}\,$ is nonzero then the above is simplified on taking $\vec{s}\,$ to be $\vec{v}$.) If we assume that $\vec{v}\,$ is nonzero then we can simplify the above by taking $\vec{s}\,$ to be $\vec{v}$.) \partsitem Write $c_{\vec{p}}\vec{s}\,$ for the projection $\proj{\vec{v}}{\spanof{\vec{s}\,}}$. Note that, by the assumption that $\vec{v}$ is not in the line, ... ... @@ -17270,7 +17270,7 @@ $(a_1+a_2)\cdot\vec{v}=a_2c_{\vec{p}}\cdot\vec{s}$. Because $\vec{v}\,$ isn't in the line, the scalars $a_1+a_2$ and $a_2 c_{\vec{p}}$ must both be zero. The $c_{\vec{p}}=0$ case is handled above, so We handled the $c_{\vec{p}}=0$ case above, so the remaining case is that $a_2=0$, and this gives that $a_1=0$ also. Hence the set is linearly independent. ... ... @@ -17357,8 +17357,8 @@ \frac{\vec{v}\dotprod\vec{s}}{\vec{s}\dotprod\vec{s}}\cdot\vec{s} \end{equation*} the distance squared from the point to the line is this (a vector dotted with itself $\vec{w}\dotprod\vec{w}$ is written $\vec{w}^2$). (we write a vector dotted with itself $\vec{w}\dotprod\vec{w}$ as $\vec{w}^2$). \begin{align*} \norm{\vec{v}- \frac{\vec{v}\dotprod\vec{s}}{\vec{s}\dotprod\vec{s}} ... ... @@ -17670,7 +17670,7 @@ \end{ans} \begin{ans}{Three.VI.2.12} The given space can be parametrized in this way. We can paramatrize the given space can in this way. \begin{equation*} \set{\colvec{x \\ y \\ z} \suchthat x=y-z} =\set{\colvec[r]{1 \\ 1 \\ 0}\cdot y+\colvec[r]{-1 \\ 0 \\ 1}\cdot z ... ... @@ -17806,10 +17806,17 @@ \end{ans} \begin{ans}{Three.VI.2.15} If that set is not linearly independent, then we get a zero vector. Otherwise (if our set is linearly independent but does not span the space), we are doing Gram-Schmidt on a set that is a basis for a subspace and so we get an orthogonal basis for a subspace. \end{ans} \begin{ans}{Three.VI.2.16} The process leaves the basis unchanged. \end{ans} \begin{ans}{Three.VI.2.16} \begin{ans}{Three.VI.2.17} \begin{exparts} \partsitem The argument is as in the $i=3$ case of the proof of \nearbytheorem{th:GramSchmidt}. ... ... @@ -17837,7 +17844,7 @@ \vec{\kappa}_i\dotprod\vec{\kappa}_i)\right) \cdot\vec{\kappa}_i$equals, after all of the cancellation is done, zero). \partsitem The vector$\vec{v}$is shown in black and the \partsitem The vector$\vec{v}$is in black and the vector$\proj{\vec{v}\,}{\spanof{\vec{\kappa}_1}} +\proj{\vec{v}\,}{\spanof{\vec{v}_2}} =1\cdot\vec{e}_1+2\cdot\vec{e}_2$is in gray. ... ... @@ -17849,15 +17856,15 @@ +\proj{\vec{v}\,}{\spanof{\vec{v}_2}})$ lies on the dotted line connecting the black vector to the gray one, that is, it is orthogonal to the $xy$-plane. \partsitem This diagram is gotten by following the hint. \partsitem We get this diagram by following the hint. \begin{center} \small \includegraphics{ch3.84} \end{center} The dashed triangle has a right angle where the gray vector $1\cdot\vec{e}_1+2\cdot\vec{e}_2$ meets the vertical dashed line $\vec{v}-(1\cdot\vec{e}_1+2\cdot\vec{e}_2)$; this is what was proved in the first item of this question. $\vec{v}-(1\cdot\vec{e}_1+2\cdot\vec{e}_2)$; this is what first item of this question proved. The Pythagorean theorem then gives that the hypoteneuse\Dash the segment from $\vec{v}$ to any other vector\Dash is longer than the vertical dashed line. ... ... @@ -17897,7 +17904,7 @@ \end{exparts} \end{ans} \begin{ans}{Three.VI.2.17} \begin{ans}{Three.VI.2.18} One way to proceed is to find a third vector so that the three together make a basis for $\Re^3$, e.g., \begin{equation*} ... ... @@ -17947,9 +17954,9 @@ including the two vectors given in the question. \end{ans} \begin{ans}{Three.VI.2.18} \begin{ans}{Three.VI.2.19} \begin{exparts} \partsitem The representation can be done by eye. \partsitem We can do the representation by eye. \begin{equation*} \colvec[r]{2 \\ 3}=3\cdot\colvec[r]{1 \\ 1}+(-1)\cdot\colvec[r]{1 \\ 0} \qquad ... ... @@ -17969,7 +17976,7 @@ \cdot\colvec[r]{1 \\ 0} =\frac{2}{1}\cdot\colvec[r]{1 \\ 0} \end{equation*} \partsitem As above, the representation can be done by eye \partsitem As above, we can do the representation by eye \begin{equation*} \colvec[r]{2 \\ 3}=(5/2)\cdot\colvec[r]{1 \\ 1} +(-1/2)\cdot\colvec[r]{1 \\ -1} ... ... @@ -18009,7 +18016,7 @@ \end{exparts} \end{ans} \begin{ans}{Three.VI.2.19} \begin{ans}{Three.VI.2.20} First, $\norm{\vec{v}\,}^2=4^2+3^2+2^2+1^2=50$. \begin{exparts*} \partsitem $c_1=4$ ... ... @@ -18085,16 +18092,16 @@ The result now follows on gathering like terms and on recognizing that $\vec{\kappa}_1\dotprod\vec{\kappa}_1=1$ and $\vec{\kappa}_2\dotprod\vec{\kappa}_2=1$ because these vectors are given as members of an orthonormal set. members of an orthonormal set. \end{ans} \begin{ans}{Three.VI.2.20} \begin{ans}{Three.VI.2.21} It is true, except for the zero vector. Every vector in $$\Re^n$$ except the zero vector is in a basis, and that basis can be orthogonalized. \end{ans} \begin{ans}{Three.VI.2.21} \begin{ans}{Three.VI.2.22} The $\nbyn{3}$ case gives the idea. The set \begin{equation*} ... ... @@ -18147,7 +18154,7 @@ \end{equation*} \end{ans} \begin{ans}{Three.VI.2.22} \begin{ans}{Three.VI.2.23} If the set is empty then the summation on the left side is the linear combination of the empty set of vectors, which by definition adds to the zero vector. ... ... @@ -18155,7 +18162,7 @@ `if \ldots then \ldots' implication is vacuously true. \end{ans} \begin{ans}{Three.VI.2.23} \begin{ans}{Three.VI.2.24} \begin{exparts} \partsitem Part of the induction argument proving \nearbytheorem{th:GramSchmidt} checks that ... ... @@ -18176,7 +18183,7 @@ \end{exparts} \end{ans} \begin{ans}{Three.VI.2.24} \begin{ans}{Three.VI.2.25} For the inductive step, we assume that for all $j$ in~$[1..i]$, these three conditions are true of each $\vec{\kappa}_j$: (i)~each $\vec{\kappa}_j$ is nonzero, ... ... @@ -18342,7 +18349,7 @@ \begin{equation*} M^\perp=\set{k\cdot\colvec[r]{1 \\ 1}\suchthat k\in\Re} \end{equation*} \partsitem As in the answer to the prior part, $M$ can be described as \partsitem As in the answer to the prior part, we can describe $M$ as a span \begin{equation*} M=\set{c\cdot\colvec[r]{3/2 \\ 1}\suchthat c\in\Re} ... ... @@ -18737,8 +18744,8 @@ \begin{ans}{Three.VI.3.14} No, a decomposition of vectors $\vec{v}=\vec{m}+\vec{n}$ into $\vec{m}\in M$ and $\vec{n}\in N$ does not depend on the bases chosen for the subspaces\Dash this was shown in the Direct Sum subsection. chosen for the subspaces, as we showed in the Direct Sum subsection. \end{ans} \begin{ans}{Three.VI.3.15} ... ... @@ -18778,7 +18785,7 @@ \end{ans} \begin{ans}{Three.VI.3.18} If $V=M\directsum N$ then every vector can be decomposed uniquely as If $V=M\directsum N$ then every vector decomposes uniquely as $\vec{v}=\vec{m}+\vec{n}$. For all $\vec{v}$ the map $p$ gives $p(\vec{v})=\vec{m}$ if and only if $\vec{v}-p(\vec{v})=\vec{n}$, as required. ... ... @@ -18980,7 +18987,7 @@ \end{equation*} we have $\nullspace{f}^\perp =\spanof{\set{\vec{h}_1,\dots,\vec{h}_m}}$. (In \cite{Strang93}, this space is described as the (\cite{Strang93} describes this space as the transpose of the row space of $H$.) \end{exparts} ... ... @@ -19035,7 +19042,7 @@ rewritten as $t(\vec{v}-t(\vec{v}))=\zero$ suggests taking $\vec{v}=t(\vec{v})+(\vec{v}-t(\vec{v}))$. So we are finished on taking a basis To finish we taking a basis $B=\sequence{\vec{\beta}_1,\ldots,\vec{\beta}_n}$ for $V$ where $\sequence{\vec{\beta}_1,\ldots,\vec{\beta}_r}$ is a basis for the rangespace $M$ and
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