Commit 0e9ce5c0 authored by Jim Hefferon's avatar Jim Hefferon

map3 answers

parent 633557a5
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......@@ -1044,8 +1044,9 @@ is more fruitful and more central to progress.
\begin{answer}
Recall that the entry in row~$i$ and column~$j$ of the transpose of $M$
is the entry $m_{j,i}$ from row~$j$ and column~$i$ of $M$.
Now, the check is routine.
\begin{multline*}
Now, the check is routine.
Start with the transpose of the combination.
\begin{equation*}
\trans{[r\cdot\begin{mat}
\ &\vdots \\
\cdots &a_{i,j} &\cdots \\
......@@ -1055,19 +1056,24 @@ is more fruitful and more central to progress.
\ &\vdots \\
\cdots &b_{i,j} &\cdots \\
&\vdots
\end{mat}]} \\
\begin{aligned}
&=\trans{\begin{mat}
\end{mat}]}
\end{equation*}
Combine and take the transpose.
\begin{equation*}
=\trans{\begin{mat}
\ &\vdots \\
\cdots &ra_{i,j}+sb_{i,j} &\cdots \\
&\vdots
\end{mat}} \\
&=\begin{mat}
\end{mat}}
=\begin{mat}
\ &\vdots \\
\cdots &ra_{j,i}+sb_{j,i} &\cdots \\
&\vdots
\end{mat} \\
&=r\cdot\begin{mat}
\end{mat}
\end{equation*}
Then bring out the scalars, and un-transpose.
\begin{multline*}
=r\cdot\begin{mat}
\ &\vdots \\
\cdots &a_{j,i} &\cdots \\
&\vdots
......@@ -1076,8 +1082,8 @@ is more fruitful and more central to progress.
\ &\vdots \\
\cdots &b_{j,i} &\cdots \\
&\vdots
\end{mat} \\
&=r\cdot\trans{\begin{mat}
\end{mat} \\
=r\cdot\trans{\begin{mat}
\ &\vdots \\
\cdots &a_{j,i} &\cdots \\
&\vdots
......@@ -1087,7 +1093,6 @@ is more fruitful and more central to progress.
\cdots &b_{j,i} &\cdots \\
&\vdots
\end{mat} }
\end{aligned}
\end{multline*}
The domain is \( \matspace_{\nbym{m}{n}} \)
while the codomain is \( \matspace_{\nbym{n}{m}} \).
......
......@@ -706,19 +706,22 @@ for any matrix there is an associated linear map.
with respect to $\stdbasis_i$, a column vector represents itself.
\begin{exparts}
\partsitem To represent \( h \) with respect
to \( \stdbasis_2,\stdbasis_3 \) we take
to \( \stdbasis_2,\stdbasis_3 \) take
the images of the basis vectors from the domain,
and represent them with respect to the basis for the codomain.
The first is this
\begin{equation*}
\rep{\,h(\vec{e}_1)\,}{\stdbasis_3}
=\rep{\colvec[r]{2 \\ 2 \\ 0}}{\stdbasis_3}
=\colvec[r]{2 \\ 2 \\ 0}
\qquad
\end{equation*}
while the second is this.
\begin{equation*}
\rep{\,h(\vec{e}_2)\,}{\stdbasis_3}
=\rep{\colvec[r]{0 \\ 1 \\ -1}}{\stdbasis_3}
=\colvec[r]{0 \\ 1 \\ -1}
\end{equation*}
These are adjoined to make the matrix.
Adjoin these to make the matrix.
\begin{equation*}
\rep{h}{\stdbasis_2,\stdbasis_3}=
\begin{mat}[r]
......@@ -761,15 +764,15 @@ for any matrix there is an associated linear map.
\partsitem
We must first find the image of each vector from the domain's basis,
and then represent that image with respect to the codomain's basis.
\begin{equation*}
\begin{multline*}
\rep{\frac{d\,1}{dx}}{B}=\colvec[r]{0 \\ 0 \\ 0 \\ 0}
\quad
\rep{\frac{d\,x}{dx}}{B}=\colvec[r]{1 \\ 0 \\ 0 \\ 0}
\quad
\rep{\frac{d\,x^2}{dx}}{B}=\colvec[r]{0 \\ 2 \\ 0 \\ 0}
\quad
\\
\rep{\frac{d\,x^3}{dx}}{B}=\colvec[r]{0 \\ 0 \\ 3 \\ 0}
\end{equation*}
\end{multline*}
Those representations are then adjoined to make the matrix
representing the map.
\begin{equation*}
......@@ -783,15 +786,15 @@ for any matrix there is an associated linear map.
\end{equation*}
\partsitem Proceeding as in the prior item, we represent the images
of the domain's basis vectors
\begin{equation*}
\begin{multline*}
\rep{\frac{d\,1}{dx}}{D}=\colvec[r]{0 \\ 0 \\ 0 \\ 0}
\quad
\rep{\frac{d\,x}{dx}}{D}=\colvec[r]{1 \\ 0 \\ 0 \\ 0}
\quad
\rep{\frac{d\,x^2}{dx}}{D}=\colvec[r]{0 \\ 1 \\ 0 \\ 0}
\quad
\\
\rep{\frac{d\,x^3}{dx}}{D}=\colvec[r]{0 \\ 0 \\ 1 \\ 0}
\end{equation*}
\end{multline*}
and adjoin to make the matrix.
\begin{equation*}
\rep{\frac{d}{dx}}{B,D}=
......@@ -861,15 +864,15 @@ for any matrix there is an associated linear map.
\end{equation*}
and these images are represented with respect to the codomain's
basis in this way.
\begin{equation*}
\begin{multline*}
\rep{0}{B}=\colvec{0 \\ 0 \\ 0 \\ \vdots \\ \ \\ \ }
\quad
\rep{1}{B}=\colvec{1 \\ 0 \\ 0 \\ \vdots \\ \ \\ \ }
\quad
\rep{2x}{B}=\colvec{0 \\ 2 \\ 0 \\ \vdots \\ \ \\ \ }
\quad\ldots\quad
\rep{2x}{B}=\colvec{0 \\ 2 \\ 0 \\ \vdots \\ \ \\ \ } \\
\ldots\quad
\rep{nx^{n-1}}{B}=\colvec{0 \\ 0 \\ 0 \\ \vdots \\ n \\ 0}
\end{equation*}
\end{multline*}
The matrix
\begin{equation*}
\rep{\frac{d}{dx}}{B,B}
......@@ -891,14 +894,14 @@ for any matrix there is an associated linear map.
\quad \ldots
\end{equation*}
then they can be represented with respect to the codomain's basis
\begin{equation*}
\begin{multline*}
\rep{x}{B_{n+1}}=\colvec{0 \\ 1 \\ 0 \\ \vdots \\ \ }
\quad
\rep{x^2/2}{B_{n+1}}=\colvec{0 \\ 0 \\ 1/2 \\ \vdots \\ \ }
\quad\ldots\quad
\rep{x^2/2}{B_{n+1}}=\colvec{0 \\ 0 \\ 1/2 \\ \vdots \\ \ } \\
\ldots\quad
\rep{x^{n+1}/(n+1)}{B_{n+1}}
=\colvec{0 \\ 0 \\ 0 \\ \vdots \\ 1/(n+1)}
\end{equation*}
\end{multline*}
and put together to make the matrix.
\begin{equation*}
\rep{\int}{B_{n},B_{n+1}}
......@@ -953,14 +956,11 @@ for any matrix there is an associated linear map.
\end{mat}
\end{equation*}
\partsitem The images of the basis vectors from the domain are
\begin{equation*}
1\mapsto 1
\quad x\mapsto x+1=1+x
\quad x^2\mapsto (x+1)^2=1+2x+x^2
\quad x^3\mapsto (x+1)^3=1+3x+3x^2+x^3
\quad \ldots
\end{equation*}
which are represented as
$1\mapsto 1$,
and $x\mapsto x+1=1+x$,
and $x^2\mapsto (x+1)^2=1+2x+x^2$,
and $x^3\mapsto (x+1)^3=1+3x+3x^2+x^3$, etc.
The representations are here.
\begin{equation*}
\rep{1}{B}=\colvec[r]{1 \\ 0 \\ 0 \\ 0 \\ \vdotswithin{0} \\ 0}
\quad
......@@ -1355,13 +1355,13 @@ for any matrix there is an associated linear map.
\end{equation*}
and those images are represented with respect to the codomain's
basis in this way.
\begin{equation*}
\begin{multline*}
\rep{\,h(\vec{\beta}_1)\,}{h(B)}=\colvec[r]{1 \\ 0 \\ \vdotswithin{0} \\ 0}
\quad
\rep{\,h(\vec{\beta}_2)\,}{h(B)}=\colvec[r]{0 \\ 1 \\ \vdotswithin{0} \\ 0}
\quad\ldots\quad
\rep{\,h(\vec{\beta}_2)\,}{h(B)}=\colvec[r]{0 \\ 1 \\ \vdotswithin{0} \\ 0} \\
\ldots\quad
\rep{\,h(\vec{\beta}_n)\,}{h(B)}=\colvec[r]{0 \\ 0 \\ \vdotswithin{0} \\ 1}
\end{equation*}
\end{multline*}
Hence, the matrix is the identity.
\begin{equation*}
\rep{h}{B,h(B)}
......@@ -1508,7 +1508,7 @@ for any matrix there is an associated linear map.
is the $y$~axis.
\partsitem The set of vectors represented with
respect to $\stdbasis_2$ as
\begin{equation*}
\begin{align*}
\set{
\begin{mat}
a &b \\
......@@ -1516,11 +1516,11 @@ for any matrix there is an associated linear map.
\end{mat}
\colvec{x \\ y}
\suchthat x,y\in\Re}
=\set{\colvec{ax+by \\ 2ax+2by}
&=\set{\colvec{ax+by \\ 2ax+2by}
\suchthat x,y\in\Re} \\
&=\set{(ax+by)\cdot\colvec[r]{1 \\ 2}
\suchthat x,y\in\Re}
=\set{(ax+by)\cdot\colvec[r]{1 \\ 2}
\suchthat x,y\in\Re}
\end{equation*}
\end{align*}
is the line $y=2x$, provided either $a$ or $b$ is not zero, and
is the set consisting of just the origin if both are zero.
\end{exparts}
......@@ -1751,17 +1751,17 @@ for any matrix there is an associated linear map.
\begin{answer}
\begin{exparts}
\partsitem
Write \( B_U \) as
Write the basis \( B_U \) as
\( \sequence{\vec{\beta}_1,\dots,\vec{\beta}_k} \) and
then $B_V$ as \( \sequence{\vec{\beta}_1,\dots,\vec{\beta}_k,
then write $B_V$ as the extension
\( \sequence{\vec{\beta}_1,\dots,\vec{\beta}_k,
\vec{\beta}_{k+1},\dots,\vec{\beta}_n} \).
If
\begin{equation*}
\rep{\vec{v}}{B_U}=\colvec{c_1 \\ \vdots \\ c_k}
\qquad\text{so that\ }
\vec{v}=c_1\cdot\vec{\beta}_1+\cdots+c_k\cdot\vec{\beta}_k
\end{equation*}
then,
so that $\vec{v}=c_1\cdot\vec{\beta}_1+\cdots+c_k\cdot\vec{\beta}_k$
then
\begin{equation*}
\rep{\vec{v}}{B_V}=\colvec{c_1 \\ \vdots\\ c_k \\ 0 \\ \vdots \\ 0}
\end{equation*}
......@@ -2855,16 +2855,16 @@ And, we shall see how to find the matrix that represents a map's inverse.
start with $\stdbasis_3$ as the basis for the domain, and then
we require a basis $D$ for the codomain $\Re^3$.
The matrix $H$ gives the action of the map as this
\begin{equation*}
\begin{multline*}
\colvec[r]{1 \\ 0 \\ 0}=\colvec[r]{1 \\ 0 \\ 0}_{\stdbasis_3}
\mapsto\colvec[r]{1 \\ 2 \\ 0}_D
\quad
\colvec[r]{0 \\ 1 \\ 0}=\colvec[r]{0 \\ 1 \\ 0}_{\stdbasis_3}
\mapsto\colvec[r]{0 \\ 0 \\ 1}_D
\quad
\\ %\quad
\colvec[r]{0 \\ 0 \\ 1}=\colvec[r]{0 \\ 0 \\ 1}_{\stdbasis_3}
\mapsto\colvec[r]{0 \\ 0 \\ 0}_D
\end{equation*}
\end{multline*}
and there is no harm in finding a basis $D$ so that
\begin{equation*}
\rep{\colvec[r]{1 \\ 0 \\ 0}}{D}=\colvec[r]{1 \\ 2 \\ 0}_D
......
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