### draft of map2 edits

parent 21c1e531
 ... ... @@ -13,6 +13,7 @@ TODO list for Linear Algebra http://joshua.smcvt.edu/linearalgebra ** subsectionoptional results in pdf errors in the log file about math shift * New material ** topic on Google's algorithm? ... ... @@ -25,6 +26,10 @@ TODO list for Linear Algebra http://joshua.smcvt.edu/linearalgebra ** Hestenes? ** Topic on magic squares? http://mathoverflow.net/questions/33911/why-linear-algebra-is-funor ** Discuss writegood mode in 1st read me. Mention that C-cG turns it off. * Refactor some code ... ... @@ -58,6 +63,8 @@ one file to read from another? That is, cannot put files by chapter? See http://tex.stackexchange.com/questions/37167/sorting-logs-into-different-folders ** Use the numprint package? http://www.ctan.org/pkg/numprint ... ...
This diff is collapsed.
 ... ... @@ -11054,8 +11054,8 @@ \end{ans} \begin{ans}{Three.II.1.18} For each, we must either check that linear combinations are preserved, or give an example of a linear combination that is For each, we must either check that the map preserves linear combinations or give an example of a linear combination that is not. \begin{exparts*} \partsitem Yes. ... ... @@ -11219,7 +11219,7 @@ 0 &1 \end{mat} \end{equation*} are mapped to the same member of the codomain, $1\in\Re$. map to the same member of the codomain, $1\in\Re$. \end{ans} \begin{ans}{Three.II.1.22} ... ... @@ -11394,8 +11394,7 @@ (This argument has already appeared, as part of the proof that isomorphism is an equivalence.) Let $\map{f}{U}{V}$ and $\map{g}{V}{W}$ be linear. For any $\vec{u}_1,\vec{u}_2\in U$ and scalars $c_1,c_2$ combinations are preserved. The composition preserves linear combinations \begin{multline*} \composed{g}{f}(c_1\vec{u}_1+c_2\vec{u}_2) =g(\,f(c_1\vec{u}_1+c_2\vec{u}_2)\,) ... ... @@ -11404,6 +11403,7 @@ =c_1\cdot \composed{g}{f}(\vec{u}_1) +c_2\cdot \composed{g}{f}(\vec{u}_2) \end{multline*} where $\vec{u}_1,\vec{u}_2\in U$ and scalars $c_1,c_2$ \end{ans} \begin{ans}{Three.II.1.33} ... ... @@ -11528,7 +11528,7 @@ Because the endpoints of $\ell$ are in the image of $C$, there are members of $C$ that map to them, say $h(\vec{c}_1)=\vec{d}_1$ and $h(\vec{c}_2)=\vec{d}_2$. Now, where $\hat{t}$ is the scalar that is fixed in the first Now, where $\hat{t}$ is the scalar that we fixed in the first sentence of this paragraph, observe that $h(\hat{t}\cdot\vec{c}_1+(1-\hat{t})\cdot\vec{c}_2) =\hat{t}\cdot h(\vec{c}_1)+(1-\hat{t})\cdot h(\vec{c}_2) ... ... @@ -11642,7 +11642,7 @@ If$\sequence{\vec{\beta}_1,\ldots,\vec{\beta}_n}$doesn't span the space then the map needn't be unique. For instance, if we try to define a map from$\Re^2$to itself by specifying only that$\vec{e}_1$is sent to itself, then specifying only that$\vec{e}_1\$ maps to itself, then there is more than one homomorphism possible; both the identity map and the projection map onto the first component fit this condition. ... ... @@ -11688,7 +11688,7 @@ \end{equation*} They are linear because they are the composition of linear functions, and the fact that the compoistion of linear functions is linear was shown as part of the proof that isomorphism is an equivalence was part of the proof that isomorphism is an equivalence relation (alternatively, the check that they are linear is straightforward). \partsitem In general, a map from a vector space $$V$$ to an
This diff is collapsed.
This diff is collapsed.
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!