Commit 2320544b authored by Jim Hefferon's avatar Jim Hefferon

change rangespace to range space

parent f3053e59
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......@@ -607,7 +607,7 @@ if and only if they have the same rank).
To prove the statement without quoting the results about
matrix equivalence, note first that
rank is a property of the map (it is the dimension of the rangespace)
rank is a property of the map (it is the dimension of the range space)
and since we've shown that
the rank of a map is the rank of a representation,
it must be the same for all representations.
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......@@ -109,18 +109,18 @@ the sum down its upper-left to lower-right diagonal
$\trace (M)=m_{1,1}+\cdots+m_{n,n}$.
Consider the restriction of the trace to the magic squares
$\map{\trace}{\magicsquares_n}{\Re}$.
The nullspace $\nullspace{\trace}$ is the set of magic squares with magic
The null space $\nullspace{\trace}$ is the set of magic squares with magic
number zero
$\magicsquares_{n,0}$.
Observe that the trace is onto because for any~$r$ in the
codomain $\Re$ the $\nbyn{n}$ matrix whose entries are all $r/n$ is
a magic square with magic number~$r$.
Theorem~Two.II.\ref{th:RankPlusNullEqDim} says that for any linear map the
dimension of the domain equals the dimension of the rangespace
plus the dimension of the nullspace,
dimension of the domain equals the dimension of the range space
plus the dimension of the null space,
the map's rank plus its nullity.
Here the domain is $\magicsquares_n$, the rangespace is
$\Re$ and the nullspace is $\magicsquares_{n,0}$,
Here the domain is $\magicsquares_n$, the range space is
$\Re$ and the null space is $\magicsquares_{n,0}$,
so we have that $\dim\magicsquares_n=1+\dim\magicsquares_{n,0}$.
We will finish by showing that
......@@ -156,7 +156,7 @@ With respect to the standard basis, each represents a linear map
$\map{h}{\Re^{n^2}}{\Re^{2n+2}}$.
The domain has dimension~$n^2$ so if we show that the
rank of the matrix is $2n+1$ then we will have
what we want, that the dimension of the nullspace
what we want, that the dimension of the null space
$\magicsquares_{n,0}$ is $n^2-(2n+1)$.
\newlength{\interblockhspace}
\setlength{\interblockhspace}{1.45em}
......@@ -434,7 +434,7 @@ The proof given here began with \cite{Ward}.
the ordered pair $(\trace(M),\trace^*(M))$.
Specifically, consider the restriction of that map
$\map{\theta}{\semimagicsquares_n}{\Re^2}$ to the semimagic squares.
Clearly its nullspace is $\magicsquares_{n,0}$.
Clearly its null space is $\magicsquares_{n,0}$.
Show that when $n\geq 3$ this restriction $\theta$ is onto.
(\textit{Hint:} we need only find a basis for
$\Re^2$ that is the
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......@@ -1772,8 +1772,8 @@ for any matrix there is an associated linear map.
We must first decide what the question means.
Compare \( \map{h}{V}{W} \) with its restriction to the subspace
\( \map{\restrictionmap{h}{U}}{U}{W} \).
The rangespace of the restriction is a subspace of \( W \), so fix a
basis \( D_{h(U)} \) for this rangespace and extend it to a basis
The range space of the restriction is a subspace of \( W \), so fix a
basis \( D_{h(U)} \) for this range space and extend it to a basis
\( D_V \) for \( W \).
We want the relationship between these two.
\begin{equation*}
......@@ -2012,7 +2012,7 @@ Suppose that the matrix \( H \) is \( \nbym{m}{n} \).
Fix domain and codomain spaces $V$ and $W$ of dimension $n$ and~$m$ with
bases \( B=\sequence{\vec{\beta}_1,\dots,\vec{\beta}_n} \) and \( D \).
Then \( H \) represents some linear map $h$ between those spaces with respect
to these bases whose rangespace
to these bases whose range space
\begin{align*}
\set{h(\vec{v})\suchthat \vec{v}\in V}
&=\set{h(c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n)
......@@ -2021,7 +2021,7 @@ to these bases whose rangespace
\suchthat c_1,\dots,c_n\in\Re}
\end{align*}
is the span $\spanof{\set{h(\vec{\beta}_1),\dots,h(\vec{\beta}_n)}}$.
The rank of the map $h$ is the dimension of this rangespace.
The rank of the map $h$ is the dimension of this range space.
The rank of the matrix is the dimension of its column space,
the span of the set of its columns
......@@ -2032,14 +2032,14 @@ from the proof of Lemma~I.\ref{lem:EqDimImpIso} that if we fix a basis then
representation with respect to that basis gives an isomorphism
$\map{\mbox{Rep}_D}{W}{\Re^m}$.
Under this isomorphism there is a linear relationship among members of the
rangespace if and only if the same relationship holds in the
range space if and only if the same relationship holds in the
column space, e.g,
$\zero=c_1h(\vec{\beta}_1)+\dots+c_nh(\vec{\beta}_n)$ if and only if
$\zero=c_1\rep{h(\vec{\beta}_1)}{D}+\dots+c_n\rep{h(\vec{\beta}_n)}{D}$.
Hence, a subset of the rangespace is linearly independent if and only if the
Hence, a subset of the range space is linearly independent if and only if the
corresponding subset of the column space is linearly independent.
Therefore the size of the largest linearly independent subset of the
rangespace equals the size of the largest linearly independent subset of the
range space equals the size of the largest linearly independent subset of the
column space, and so the two spaces have the same dimension.
\end{proof}
......@@ -2057,7 +2057,7 @@ must be from a three-dimensional domain
to a four-dimensional codomain.
In addition, because the rank of this matrix is two
(we can spot this by eye or get it with Gauss' method),
any map represented by this matrix has a two-dimensional rangespace.
any map represented by this matrix has a two-dimensional range space.
\end{example}
\begin{corollary} \label{cor:MatDescsMap}
......@@ -2069,17 +2069,17 @@ its columns.
\end{corollary}
\begin{proof}
For the onto half, the dimension of the rangespace of $h$ is the rank of $h$,
For the onto half, the dimension of the range space of $h$ is the rank of $h$,
which equals the rank of $H$ by the theorem.
Since the dimension of the codomain of $h$ equals the number of rows in $H$,
if the rank of $H$ equals the number of rows then the dimension of the
rangespace equals the dimension of the codomain.
range space equals the dimension of the codomain.
But a subspace with the same dimension as its superspace must equal
that superspace
(because any basis for the rangespace is a linearly independent subset
(because any basis for the range space is a linearly independent subset
of the codomain
whose size is equal to the dimension of the codomain, and thus so this
basis for the rangespace must also be
basis for the range space must also be
a basis for the codomain).
For the other half,
......@@ -2107,7 +2107,7 @@ Some authors use `nonsingular' as a synonym for one-to-one
while others use it the way that we have here, as a synonym for
isomorphism.
The difference is slight because a one-to-one map is onto its
rangespace.
range space.
\end{remark}
In the first chapter we defined a matrix to be nonsingular
......@@ -2498,10 +2498,10 @@ And, we shall see how to find the matrix that represents a map's inverse.
nonsingular and is therefore associated with maps that are nonsingular.
\begin{exparts}
\partsitem Find the set of column vectors representing the members of
the nullspace of any map represented by this matrix.
the null space of any map represented by this matrix.
\partsitem Find the nullity of any such map.
\partsitem Find the set of column vectors representing the members of
the rangespace of any map represented by this matrix.
the range space of any map represented by this matrix.
\partsitem Find the rank of any such map.
\partsitem Check that rank plus nullity equals the dimension of the
domain.
......@@ -2523,7 +2523,7 @@ And, we shall see how to find the matrix that represents a map's inverse.
\begin{equation*}
\rep{\zero}{D}=\colvec[r]{0 \\ 0}_D
\end{equation*}
and so the set of representations of members of the nullspace is
and so the set of representations of members of the null space is
this.
\begin{equation*}
\set{\colvec{x \\ y}_B\suchthat \text{$x+2y=0$ and $3x+6y=0$}}
......@@ -2535,9 +2535,9 @@ And, we shall see how to find the matrix that represents a map's inverse.
The subspace of $\Re^2$ that is in the prior item is
one-dimensional.
Therefore, the image of that subspace under the inverse of the
representation map\Dash the nullspace of $G$,
representation map\Dash the null space of $G$,
is also one-dimensional.
\partsitem The set of representations of members of the rangespace is
\partsitem The set of representations of members of the range space is
this.
\begin{equation*}
\set{\colvec{x+2y \\ 3x+6y}_D\suchthat x,y\in\Re}
......@@ -2546,7 +2546,7 @@ And, we shall see how to find the matrix that represents a map's inverse.
\partsitem Of course, \nearbytheorem{th:RankMatEqRankMap} gives that
the rank of the map equals the rank of the matrix, which is one.
Alternatively, the same argument that we used above for the
nullspace gives here that the dimension of the rangespace is one.
null space gives here that the dimension of the range space is one.
\partsitem One plus one equals two.
\end{exparts}
\end{answer}
......@@ -2556,11 +2556,11 @@ And, we shall see how to find the matrix that represents a map's inverse.
one matrix represents two different maps
\( H=\rep{h}{B,D}=\rep{\hat{h}}{\hat{B},\hat{D}} \)
(where \( \map{h,\hat{h}}{V}{W} \))
then the dimension of the rangespace of
\( h \) equals the dimension of the rangespace of \( \hat{h} \).
Must these equal-dimensioned rangespaces actually be the same?
then the dimension of the range space of
\( h \) equals the dimension of the range space of \( \hat{h} \).
Must these equal-dimensioned range spaces actually be the same?
\begin{answer}
No, the rangespaces may differ.
No, the range spaces may differ.
\nearbyexample{ex:CngBasesChgMap} shows this.
\end{answer}
\recommended \item
......
......@@ -2273,7 +2273,7 @@ We can express those conditions more compactly as a linear system.
\colvec{v_1 \\ v_2 \\ v_3}
=\colvec[r]{0 \\ 0} }
\end{equation*}
We are thus left with finding the nullspace of the map represented
We are thus left with finding the null space of the map represented
by the matrix, that is, with
calculating the solution set of a homogeneous linear system.
\begin{equation*}
......@@ -2337,7 +2337,7 @@ vector in $M$.
\begin{proof}
First, the orthogonal complement $M^\perp$ is a subspace of $\Re^n$ because,
as noted in the prior two examples, it is a nullspace.
as noted in the prior two examples, it is a null space.
Next,
start with any basis \( B_M=\sequence{\vec{\mu}_1,\dots,\vec{\mu}_k} \)
......@@ -3227,8 +3227,8 @@ Note, as a check, that this result is indeed in $P$.
\end{answer}
\recommended \item
The material in this subsection allows us to express a geometric
relationship that we have not yet seen between the rangespace and the
nullspace of a linear map.
relationship that we have not yet seen between the range space and the
null space of a linear map.
\begin{exparts}
\partsitem Represent $\map{f}{\Re^3}{\Re}$ given by
\begin{equation*}
......@@ -3240,7 +3240,7 @@ Note, as a check, that this result is indeed in $P$.
\begin{equation*}
\colvec[r]{1 \\ 2 \\ 3}
\end{equation*}
is a member of the perp of the nullspace.
is a member of the perp of the null space.
Prove that $\nullspace{f}^\perp$ is equal to the span of this
vector.
\partsitem Generalize that to apply to any $\map{f}{\Re^n}{\Re}$.
......@@ -3256,7 +3256,7 @@ Note, as a check, that this result is indeed in $P$.
\colvec[r]{1 \\ 2 \\ 3},
\;\colvec[r]{4 \\ 5 \\ 6}
\end{equation*}
are both members of the perp of the nullspace.
are both members of the perp of the null space.
Prove that $\nullspace{f}^\perp$ is the span of these two.
(\textit{Hint}.
See the third item of \nearbyexercise{exer:AlgOfPerps}.)
......@@ -3317,7 +3317,7 @@ Note, as a check, that this result is indeed in $P$.
4 &5 &6
\end{mat}
\end{equation*}
and so the nullspace is this set.
and so the null space is this set.
\begin{equation*}
\nullspace{f}
\set{\colvec{v_1 \\ v_2 \\ v_3}
......@@ -3336,7 +3336,7 @@ Note, as a check, that this result is indeed in $P$.
\end{equation*}
and since $\nullspace{f}^\perp$ is a subspace of $\Re^n$,
the span of the two vectors is a subspace of the perp of
the nullspace.
the null space.
To see that this containment is an equality, take
\begin{equation*}
M=\spanof{\set{\colvec[r]{1 \\ 2 \\ 3}}}
......@@ -3356,7 +3356,7 @@ Note, as a check, that this result is indeed in $P$.
\vdots \\
h_{m,1}v_1+h_{m,2}v_2+\dots+h_{m,n}v_n}
\end{equation*}
and the description of the nullspace gives that
and the description of the null space gives that
on transposing the $m$~rows of $H$
\begin{equation*}
\vec{h}_1=\colvec{h_{1,1} \\ h_{1,2} \\ \vdots \\ h_{1,n}},
......@@ -3427,25 +3427,25 @@ Note, as a check, that this result is indeed in $P$.
Thus, projecting twice into $M$ along $N$ has the same effect as
projecting once.
\partsitem As suggested by the prior items, the condition
gives that $t$ leaves vectors in the rangespace unchanged,
gives that $t$ leaves vectors in the range space unchanged,
and hints that we should take
$\vec{\beta}_1$, \ldots, $\vec{\beta}_r$
to be basis vectors for the range, that is, that we should take
the range space of $t$ for $M$ (so that $\dim(M)=r$).
As for the complement, we write $N$ for the nullspace of $t$
As for the complement, we write $N$ for the null space of $t$
and we will show that $V=M\directsum N$.
To show this, we can show that their intersection
is trivial $M\intersection N=\set{\zero}$ and that they sum to
the entire space $M+N=V$.
For the first, if a vector $\vec{m}$ is in the rangespace
For the first, if a vector $\vec{m}$ is in the range space
then there is a $\vec{v}\in V$ with $t(\vec{v})=\vec{m}$,
and the condition on $t$ gives that
$t(\vec{m})=(\composed{t}{t})\,(\vec{v})=t(\vec{v})=\vec{m}$, while
if that same vector is also in the nullspace then $t(\vec{m})=\zero$
and so the intersection of the rangespace and nullspace is trivial.
if that same vector is also in the null space then $t(\vec{m})=\zero$
and so the intersection of the range space and null space is trivial.
For the second, to write an arbitrary $\vec{v}$ as the sum of
a vector from the rangespace and a vector from the nullspace,
a vector from the range space and a vector from the null space,
the fact that the condition $t(\vec{v})=t(t(\vec{v}))$ can be
rewritten as $t(\vec{v}-t(\vec{v}))=\zero$ suggests taking
$\vec{v}=t(\vec{v})+(\vec{v}-t(\vec{v}))$.
......@@ -3453,11 +3453,11 @@ Note, as a check, that this result is indeed in $P$.
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
the range space $M$ and
$\sequence{\vec{\beta}_{r+1},\ldots,\vec{\beta}_n}$ is a
basis for the nullspace $N$.
basis for the null space $N$.
\partsitem Every projection (as defined in this exercise) is
a projection into its rangespace and along its nullspace.
a projection into its range space and along its null space.
\partsitem This also follows immediately from the third item.
\end{exparts}
\end{answer}
......
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