Commit 3b30f02f authored by Jim Hefferon's avatar Jim Hefferon

add computation exercises

parent c33cf6c5
This diff is collapsed.
......@@ -25,7 +25,7 @@ Verify that each map is a homomorphism.
ax^2+bx+c\mapsto \colvec{a+b \\ a+c}
\end{equation*}
This verifies that it preserves linear combinations.
This verifies that the map preserves linear combinations.
\begin{align*}
h(\,d_1(a_1x^2+b_1x+c_1)+d_2(a_2x^2+b_2x+c_2)\,)
&=h((d_1a_1+d_2a_2)x^2+(d_1b_1+d_2b_2)x+(d_1c_1+d_2c_2)) \\
......
......@@ -676,6 +676,93 @@ partitions the collection of vector spaces.
&=r\cdot f_1(c_1x+c_2y+c_3z)
\end{align*}
\end{answer}
\item Verify that this map is an isomorphism:
$\map{h}{\Re^4}{\matspace_{\nbyn{2}}}$ given by
\begin{equation*}
\colvec{a \\ b \\ c \\ d}
\mapsto
\begin{mat}
c &a+d \\
b &d
\end{mat}
\end{equation*}
\begin{answer}
We first verify that $h$ is one-to-one.
To do this we will show that $h(\vec{v}_1)=h(\vec{v}_2)$ implies that
$\vec{v}_1=\vec{v}_2$.
So assume that
\begin{equation*}
h(\vec{v}_1)
=
h(\colvec{a_1 \\ b_1 \\ c_1 \\ d_1})
=
h(\colvec{a_2 \\ b_2 \\ c_2 \\ d_2})
=h(\vec{v}_2)
\end{equation*}
which gives
\begin{equation*}
\begin{mat}
c_1 &a_1+d_1 \\
b_1 &d_1
\end{mat}
=
\begin{mat}
c_2 &a_2+d_2 \\
b_2 &d_2
\end{mat}
\end{equation*}
from which we conclude that
$c_1=c_2$ (by the upper-left entries),
$b_1=b_2$ (by the lower-left entries),
$d_1=d_2$ (by the lower-right entries),
and with this last we get $a_1=a_2$ (by the upper right).
Therefore $\vec{v}_1=\vec{v}_2$.
Next we will show that the map is onto, that every member of the codomain
$\matspace_{\nbyn{2}}$ is the image of some four-tall member of the domain.
So, given
\begin{equation*}
\vec{w}=
\begin{mat}
m &n \\
p &q
\end{mat}
\in\matspace_{\nbyn{2}}
\end{equation*}
observe that it is the image of this domain vector.
\begin{equation*}
\vec{v}=
\colvec{n-q \\ p \\ m \\ q}
\end{equation*}
To finish we verify that the map preserves linear combinations.
By \nearbylemma{le:PresStructIffPresCombos} this will show that the
map preserves the operations.
\begin{align*}
h(r_1\cdot\colvec{a_1 \\ b_1 \\ c_1 \\ d_1}
+
r_2\cdot\colvec{a_2 \\ b_2 \\ c_2 \\ d_2})
&=h(\colvec{r_1a_1+r_2a_2 \\ r_1b_1+r_2b_2 \\ r_1c_1+r_2c_2 \\ r_1d_1+r_2d_2}) \\
&=
\begin{mat}
r_1c_1+r_2c_2 &(r_1a_1+r_2a_2)+(r_1d_1+r_2d_2) \\
r_1b_1+r_2b_2 &r_1d_1+r_2d_2
\end{mat} \\
&=
r_1\begin{mat}
c_1 &a_1+d_1 \\
b_1 &d_1
\end{mat}
+
r_2\begin{mat}
c_2 &a_2+d_2 \\
b_2 &d_2
\end{mat} \\
&=r_1\cdot h(\colvec{a_1 \\ b_1 \\ c_1 \\ d_1})
+
r_2\cdot h(\colvec{a_2 \\ b_2 \\ c_2 \\ d_2})
\end{align*}
\end{answer}
\recommended \item
Decide whether each map is an isomorphism
(if it is an isomorphism then prove it and if it isn't then
......
......@@ -613,7 +613,8 @@ is more fruitful and more central to progress.
\end{exparts*}
\end{answer}
\recommended \item
Show that these two maps are homomorphisms.
Show that these are homomorphisms.
Are they inverse to each other?
\begin{exparts}
\partsitem \( \map{d/dx}{\polyspace_3}{\polyspace_2} \)
given by \( a_0+a_1x+a_2x^2+a_3x^3 \) maps to
......@@ -621,7 +622,6 @@ is more fruitful and more central to progress.
\partsitem \( \map{\int}{\polyspace_2}{\polyspace_3} \) given by
\( b_0+b_1x+b_2x^2 \) maps to \( b_0x+(b_1/2)x^2+(b_2/3)x^3 \)
\end{exparts}
Are these maps inverse to each other?
\begin{answer}
The check that each is a homomorphisms is routine.
Here is the check for the differentiation map.
......@@ -681,6 +681,51 @@ is more fruitful and more central to progress.
Verification that each is a homomorphism is straightforward.
(The last one, of course, is the zero transformation on $\Re^3$.)
\end{answer}
\item Verify that each map is a homomorphism.
\begin{exparts}
\partsitem $\map{h}{\polyspace_3}{\Re^2}$ given by
\begin{equation*}
ax^2+bx+c\mapsto \colvec{a+b \\ a+c}
\end{equation*}
\partsitem $\map{f}{\Re^2}{\Re^3}$ given by
\begin{equation*}
\colvec{x \\ y}\mapsto\colvec{0 \\ x-y \\ 3y}
\end{equation*}
\end{exparts}
\begin{answer}
\begin{exparts}
\partsitem
This verifies that the map preserves linear combinations.
By \nearbylemma{le:HomoPreserveLinCombo} that suffices to show that
it is a homomorphism.
\begin{multline*}
h(\,d_1(a_1x^2+b_1x+c_1)+d_2(a_2x^2+b_2x+c_2)\,) \\
\begin{aligned}
&=h((d_1a_1+d_2a_2)x^2+(d_1b_1+d_2b_2)x+(d_1c_1+d_2c_2)) \\
&=\colvec{(d_1a_1+d_2a_2)+(d_1b_1+d_2b_2) \\
(d_1a_1+d_2a_2)+(d_1c_1+d_2c_2)} \\
&=\colvec{d_1a_1+d_1b_1 \\ d_1a_1+d_1c_1}+
\colvec{d_2a_2+d_2b_2 \\ d_2a_2+d_2c_2} \\
&=d_1\colvec{a_1+b_1 \\ a_1+c_1}+
d_2\colvec{a_2+b_2 \\ a_2+c_2} \\
&=d_1\cdot h(a_1x^2+b_1x+c_1)+d_2\cdot h(a_2x^2+b_2x+c_2)
\end{aligned}
\end{multline*}
\partsitem
It preserves linear combinations.
\begin{align*}
f(\,a_1\colvec{x_1 \\ y_1}+a_2\colvec{x_2 \\ y_2}\,)
&=f(\,\colvec{a_1x_1+a_2x_2 \\ a_1y_1+a_2y_2}\,) \\
&=\colvec{0 \\ (a_1x_1+a_2x_2)-(a_1y_1+a_2y_2) \\
3(a_1y_1+a_2y_2) } \\
&=a_1\colvec{0 \\ x_1-y_1 \\ 3y_1}
+a_2\colvec{0 \\ x_2-y_2 \\ 3y_2} \\
&=a_1f(\colvec{x_1 \\ y_1})+a_2f(\colvec{x_2 \\ y_2})
\end{align*}
\end{exparts}
\end{answer}
\item
Show that, while the maps from \nearbyexample{exam:TwoMapsHomoNotIso}
preserve linear operations, they are not isomorphisms.
......@@ -2097,8 +2142,41 @@ point and the map is an isomorphism between the domain and the range.
range space because of the constant term.
\end{exparts}
\end{answer}
\item Find the range space and the rank of each homomorphism.
\begin{exparts}
\partsitem
$\map{h}{\polyspace_3}{\Re^2}$ given by
\begin{equation*}
ax^2+bx+c\mapsto \colvec{a+b \\ a+c}
\end{equation*}
\partsitem $\map{f}{\Re^2}{\Re^3}$ given by
\begin{equation*}
\colvec{x \\ y}\mapsto\colvec{0 \\ x-y \\ 3y}
\end{equation*}
\end{exparts}
\begin{answer}
\begin{exparts}
\partsitem
The range of $h$ is all of the codomain~$\Re^2$ because given
\begin{equation*}
\colvec{x \\ y}\in\Re^2
\end{equation*}
it is the image under~$h$ of the domain vector $0x^2+bx+c$.
So the rank of~$h$ is~$2$.
\partsitem
The range is the $yz$~plane. Any
\begin{equation*}
\colvec{0 \\ a \\ b}
\end{equation*}
is the image under~$f$ of this domain vector.
\begin{equation*}
\colvec{a+b/3 \\ b/3}
\end{equation*}
So the rank of the map is~$2$.
\end{exparts}
\end{answer}
\recommended \item
Find the null space, nullity, range space, and rank of each map.
Find the range space and rank of each map.
\begin{exparts}
\partsitem \( \map{h}{\Re^2}{\polyspace_3} \) given by
\begin{equation*}
......@@ -2124,20 +2202,40 @@ point and the map is an isomorphism between the domain and the range.
\end{exparts}
\begin{answer}
\begin{exparts}
\partsitem The null space is
\begin{equation*}
\nullspace{h}
=\set{\colvec{a \\ b}\in\Re^2\suchthat
a+ax+ax^2+0x^3=0+0x+0x^2+0x^3}
=\set{\colvec{0 \\ b}\suchthat b\in\Re}
\end{equation*}
while the range space is
\partsitem The range space is
\begin{equation*}
\rangespace{h}
=\set{a+ax+ax^2\in\polyspace_3\suchthat a,b\in\Re}
=\set{a\cdot(1+x+x^2)\suchthat a\in\Re}
\end{equation*}
and so the nullity is one and the rank is one.
and so the rank is one.
\partsitem The range space
\begin{equation*}
\rangespace{h}=
\set{a+d\suchthat a,b,c,d\in\Re}
\end{equation*}
is all of $\Re$ (we can get any real number by
taking $d$ to be $0$ and taking $a$ to be the desired number).
Thus, the rank is one.
\partsitem The range space is
$\rangespace{h}=\set{r+sx^2\suchthat r,s\in\Re}$.
The rank is two.
\partsitem The range space is the trivial subspace of \( \Re^4 \)
so the rank is zero.
\end{exparts}
\end{answer}
\recommended \item
For each linear map in the prior exercise, find the null space and nullity.
\begin{answer}
\begin{exparts}
\partsitem The null space is
\begin{align*}
\nullspace{h}
&=\set{\colvec{a \\ b}\in\Re^2\suchthat
a+ax+ax^2+0x^3=0+0x+0x^2+0x^3} \\
&=\set{\colvec{0 \\ b}\suchthat b\in\Re}
\end{align*}
and so the nullity is one.
\partsitem The null space is this.
\begin{equation*}
\nullspace{h}
......@@ -2150,37 +2248,16 @@ point and the map is an isomorphism between the domain and the range.
c &d
\end{mat} \suchthat b,c,d\in\Re }
\end{equation*}
The range space
\begin{equation*}
\rangespace{h}=
\set{a+d\suchthat a,b,c,d\in\Re}
\end{equation*}
is all of $\Re$ (we can get any real number by
taking $d$ to be $0$ and taking $a$ to be the desired number).
Thus, the nullity is three and the rank is one.
\partsitem The null space is
\begin{equation*}
\nullspace{h}=\set{\begin{mat}
a &b \\
c &d
\end{mat} \suchthat
\text{$a+b+c=0$ and $d=0$}}
=\set{\begin{mat}
-b-c &b \\
c &0
\end{mat} \suchthat b,c\in\Re }
\end{equation*}
while the range space is
Thus, the nullity is three.
\partsitem The range space is
$\rangespace{h}=\set{r+sx^2\suchthat r,s\in\Re}$.
Thus, the nullity is two and the rank is two.
\partsitem The null space is all of \( \Re^3 \)
so the nullity is three.
The range space is the trivial subspace of \( \Re^4 \) so the rank
is zero.
Thus, the rank is two.
\partsitem The range space is the trivial subspace of \( \Re^4 \)
so the rank is zero.
\end{exparts}
\end{answer}
\recommended \item
Find the nullity of each map.
Find the nullity of each map below.
\begin{exparts*}
\partsitem \( \map{h}{\Re^5}{\Re^8} \) of rank five
\partsitem \( \map{h}{\polyspace_3}{\polyspace_3} \) of rank one
......
......@@ -2602,6 +2602,208 @@ And, we shall see how to find the matrix that represents a map's inverse.
\partsitem One plus one equals two.
\end{exparts}
\end{answer}
\recommended \item
Assume that each matrix represents a map $\map{h}{\Re^m}{\Re^n}$
with respect to the standard bases.
In each case, (i)~state $m$ and~$n$.
Then set up an augmented matrix with the given matrix on the left and
a vector representing a range space element on the right
(e.g., if the codomain is~$\Re^3$ then on the right put
$a$, $b$, and $c$).
Perform Gauss-Jordan reduction.
Use that to
(ii)~find $\rangespace{h}$ and $\rank(h)$ (and state whether the
underlying map is onto),
and (iii)~find $\nullspace{h}$ and $\nullity(h)$
(and state whether the underlying map is one-to-one).
\begin{exparts}
\partsitem $
\begin{mat}
2 &1 \\
-1 &3
\end{mat}$
\partsitem
$\begin{mat}
0 &1 &3 \\
2 &3 &4 \\
-2 &-1 &2
\end{mat}$
\partsitem
$\begin{mat}
1 &1 \\
2 &1 \\
3 &1
\end{mat}$
\partsitem Verify that the map represented by this matrix
is an isomorphism.
\begin{equation*}
\begin{mat}
2 &1 &0 \\
3 &1 &1 \\
7 &2 &1
\end{mat}
\end{equation*}
\end{exparts}
\begin{answer}
\begin{exparts}
\partsitem
(i)~The dimension of the domain space is the number of columns~$m=2$.
The dimension of the codomain space is the number of rows~$n=2$.
For the rest, we consider this matrix-vector equation.
\begin{equation*}
\begin{mat}
2 &1 \\
-1 &3
\end{mat}
\colvec{x \\ y}
=
\colvec{a \\ b}
\tag{$*$}
\end{equation*}
We solve for $x$ and~$y$.
\begin{equation*}
\begin{amat}{2}
2 &1 &a \\
-1 &3 &b
\end{amat}
\grstep{(1/2)\rho_1+\rho_2}
\grstep[(2/7)\rho_2]{(1/2)\rho_1}
\grstep{-(1/2)\rho_2+\rho_1}
\begin{amat}{2}
1 &0 &(3/7)a-(1/7)b \\
0 &1 &(1/7)a+(2/7)b
\end{amat}
\end{equation*}
(ii)~For all
\begin{equation*}
\colvec{a \\ b}\in\Re^2
\end{equation*}
in equation~($*$) the system has a solution, by the calculation.
So the range space is all of the codomain~$\rangespace{h}=\Re^2$.
The map's rank is the dimension of the range,~$2$.
The map is onto because the range space is all of the codomain.
(iii)~Again by the calculation, to find the nullspace,
setting $a=b=0$ in equation~($*$)
gives that $x=y=0$.
The null space is the trivial subspace of the domain.
\begin{equation*}
\nullspace{h}=\set{\colvec{0 \\ 0}}
\end{equation*}
The nullity is the dimension of that null space,~$0$.
The map is one-to-one because the null space is trivial.
\partsitem
(i)~The dimension of the domain space is the number of
matrix columns,~$m=3$,
and the dimension of the codomain space is the number of
rows,~$n=3$.
The calculation is this.
\begin{multline*}
\begin{amat}{3}
0 &1 &3 &a \\
2 &3 &4 &b \\
-2 &-1 &2 &c
\end{amat}
\grstep{\rho_1\leftrightarrow\rho_2}
\grstep{\rho_1+\rho_3}
\grstep{-2\rho_2+\rho_3} \\
\grstep{(1/2)\rho_1}
\grstep{-(3/2)\rho_2+\rho_1}
\begin{amat}{3}
1 &0 &-5/2 &-(3/2)a+(1/2)b \\
0 &1 &3 &a \\
0 &0 &0 &-2a+b+c
\end{amat}
\end{multline*}
(ii)~There are codomain triples
\begin{equation*}
\colvec{a \\ b \\ c}\in\Re^3
\end{equation*}
for which the system does not have a solution,
specifically the system only has a solution if $-2a+b+c=0$.
\begin{equation*}
\rangespace{h}=\set{\colvec{a \\ b \\ c}\suchthat a=(b+c)/2}
=\set{\colvec{1/2 \\ 1 \\ 0}b
+\colvec{1/2 \\ 0 \\ 1}c\suchthat b,c\in\Re}
\end{equation*}
The map's rank is the range's dimension,~$2$.
The map is not onto because the range space is not all of the
codomain.
(iii)~Setting $a=b=c=0$ in the calculation
gives infinitely many solutions.
Paramatrizing using the free variable~$z$ leads to this description
of the nullspace.
\begin{equation*}
\nullspace{h}=\set{\colvec{x \\ y \\ z}\suchthat
\text{$y=-3z$ and $x=(5/2)z$}}
=\set{\colvec{5/2 \\ -3 \\ 1}z\suchthat z\in\Re}
\end{equation*}
The nullity is the dimension of that null space,~$1$.
The map is not one-to-one because the null space is not trivial.
\partsitem
(i)~The domain has dimension $m=2$ while the codomain has
dimension~$n=3$.
Here is the calculation.
\begin{equation*}
\begin{amat}{2}
1 &1 &a \\
2 &1 &b \\
3 &1 &c
\end{amat}
\grstep[-3\rho_1+\rho_3]{-2\rho_1+\rho_2}
\grstep{-2\rho_2+\rho_3}
\grstep{-\rho_2}
\grstep{-\rho_2+\rho_1}
\begin{amat}{2}
1 &0 &-a+b \\
0 &1 &2a-b \\
0 &0 &a-2b+c
\end{amat}
\end{equation*}
(ii)~The range is this subspace of the codomain.
\begin{equation*}
\rangespace{h}=\set{\colvec{2b-c \\ b \\ c}\suchthat b,c\in\Re}
=\set{\colvec{2 \\ 1 \\ 0}b+\colvec{-1 \\ 0 \\ 1}c\suchthat b,c\in\Re}
\end{equation*}
The rank is~$2$.
The map is not onto.
(iii)~The null space is the trivial subspace of the domain.
\begin{equation*}
\nullspace{h}=\set{\colvec{x \\ y}=\colvec{0 \\ 0}}
\end{equation*}
The nullity is~$0$.
The map is one-to-one.
\partsitem Here, (i)~the domain and codomain are each of dimension~$3$.
To show (ii) and~(iii), that the map is an isomorphism,
we must show it is both onto and one-to-one.
For that we don't need to augment the matrix with $a$, $b$,
and~$c$; this calculation
\begin{multline*}
\begin{mat}
2 &1 &0 \\
3 &1 &1 \\
7 &2 &1
\end{mat}
\grstep[-(7/2)\rho_1+\rho_3]{-(3/2)\rho_1+\rho_2}
\grstep{-3\rho_2+\rho_3}
\grstep[-2\rho_2 \\ -(1/2)\rho_3]{(1/2)\rho_1}
\grstep{2\rho_3+\rho_2} \\
\grstep{-(1/2)\rho_2+\rho_1}
\begin{mat}
1 &0 &0 \\
0 &1 &0 \\
0 &0 &1
\end{mat}
\end{multline*}
gives that for each codomain vector there is one and only one
associated domain vector.
\end{exparts}
\end{answer}
\item This is an alternative proof of
\nearbylemma{le:NonsingMatIffNonsingMap}.
Given an $\nbyn{n}$ matrix $H$,
......
......@@ -1396,6 +1396,179 @@ explore this operation.
and with this, the substitution is
$\vec{d}=H\vec{x}=H(G\vec{y})=(HG)\vec{y}$.
\end{answer}
\recommended \item Consider the two linear functions
$\map{h}{\Re^3}{\polyspace_2}$
and
$\map{g}{\polyspace_2}{\matspace_{\nbyn{2}}}$
given as here.
\begin{equation*}
\colvec{a \\ b \\ c}\mapsto (a+b)x^2+(2a+2b)x+c
\qquad
px^2+qx+r\mapsto
\begin{mat}
p &p-2q \\
q &0
\end{mat}
\end{equation*}
Use these bases for the spaces.
\begin{gather*}
B=\sequence{\colvec{1 \\ 1 \\ 1},
\colvec{0 \\ 1 \\ 1},
\colvec{0 \\ 0 \\ 1}}
\qquad
C=\sequence{1+x,1-x,x^2}
\\
D=\sequence{
\begin{mat}
1 &0 \\
0 &0
\end{mat},
\begin{mat}
0 &2 \\
0 &0
\end{mat},
\begin{mat}
0 &0 \\
3 &0
\end{mat},
\begin{mat}
0 &0 \\
0 &4
\end{mat}}
\end{gather*}
\begin{exparts}
\partsitem Give the formula for the composition map
$\map{\composed{g}{h}}{\Re^3}{\matspace_{\nbyn{2}}}$ derived
directly from the above definition.
\partsitem Represent $h$ and~$g$ with respect to the appropriate bases.
\partsitem Represent the map $\composed{g}{h}$ computed in the first part
with respect to the appropriate bases.
\partsitem Check that the product of the two matrices from the second
part is the matrix from the third part.
\end{exparts}
\begin{answer}
\begin{exparts}
\partsitem
Following the definitions gives this.
\begin{align*}
\colvec{a \\ b \\ c}
&\mapsto
(a+b)x^2+(2a+2b)x+c \\
&\mapsto
\begin{mat}
a+b &(a+b)-2(2a+2b) \\
2a+2b &0
\end{mat}
=
\begin{mat}
a+b &-3a-3b \\
2a+2b &0
\end{mat}
\end{align*}
\partsitem
Because
\begin{equation*}
\colvec{1 \\ 1 \\ 1}\mapsto 2x^2+4x+1
\quad
\colvec{0 \\ 1 \\ 1}\mapsto x^2+2x+1
\quad
\colvec{0 \\ 0 \\ 1}\mapsto 0x^2+0x+1
\end{equation*}
we get this representation for~$h$.
\begin{equation*}
\rep{h}{B,C}
=
\begin{mat}
5/2 &3/2 &1/2 \\
-3/2 &-1/2 &1/2 \\
2 &1 &0
\end{mat}
\end{equation*}
Similarly, because
\begin{equation*}
1+x\mapsto
\begin{mat}
0 &-2 \\
1 &0
\end{mat}
\quad
1-x\mapsto
\begin{mat}
0 &2 \\
-1 &0
\end{mat}
\quad
x^2\mapsto
\begin{mat}
1 &1 \\
0 &0
\end{mat}
\end{equation*}
this is the representation of~$g$.
\begin{equation*}
\rep{g}{C,D}=
\begin{mat}
0 &0 &1 \\
-1 &1 &1/2 \\
1/3 &-1/3 &0 \\
0 &0 &0
\end{mat}
\end{equation*}
\partsitem
The action of $\composed{g}{h}$ on the domain basis is this.
\begin{equation*}
\colvec{1 \\ 1 \\ 1}\mapsto
\begin{mat}
2 &-6 \\
4 &0
\end{mat}
\quad
\colvec{0 \\ 1 \\ 1}\mapsto
\begin{mat}
1 &-3 \\
2 &0
\end{mat}
\quad
\colvec{0 \\ 0 \\ 1}\mapsto
\begin{mat}
0 &0 \\
0 &0
\end{mat}
\end{equation*}
We have this.
\begin{equation*}
\rep{\composed{g}{h}}{B,D}=
\begin{mat}
2 &1 &0 \\
-3 &-3/2 &0 \\
4/3 &2/3 &0 \\
0 &0 &0
\end{mat}
\end{equation*}
\partsitem
The matrix multiplication is routine, just take care with the order.
\begin{equation*}
\begin{mat}
0 &0 &1 \\
-1 &1 &1/2 \\
1/3 &-1/3 &0 \\
0 &0 &0
\end{mat}
\begin{mat}
5/2 &3/2 &1/2 \\
-3/2 &-1/2 &1/2 \\
2 &1 &0
\end{mat}
=
\begin{mat}
2 &1 &0 \\
-3 &-3/2 &0 \\
4/3 &2/3 &0 \\
0 &0 &0
\end{mat}
\end{equation*}
\end{exparts}
\end{answer}
\item
As \nearbydefinition{def:MatMult} points out, the matrix product
operation generalizes the dot product.
......@@ -1517,7 +1690,7 @@ explore this operation.
It would not represent linear map composition;
\nearbytheorem{th:MatMultRepComp} would fail.
\end{answer}
\recommended \item \label{exer:NicePropsMatMult}
\item \label{exer:NicePropsMatMult}
\begin{exparts}
\partsitem Prove that $H^pH^q=H^{p+q}$ and $(H^p)^q=H^{pq}$
for positive integers \( p,q \).
......
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