diff --git a/exercises/axler-3d.md b/exercises/axler-3d.md index 4dee591..bf142a4 100644 --- a/exercises/axler-3d.md +++ b/exercises/axler-3d.md @@ -73,7 +73,7 @@ stateDiagram-v2 The “if” direction is the most straighforward. First, if such an $E$ exists, then $\text{null }S \subset \text{null }T$. (To see this, suppose $v\in\text{null }T$; that is, $T(v)=\mathbf{0}$. Since $S(v) = -E(T(v))$ we have $S(v)=\mathbf{0}$ and hence $v\in\text{null S}$.) Now +E(T(v))$ we have $S(v)=E(\mathbf{0}) = \mathbf{0}$ and hence $v\in\text{null S}$.) Now run the argument again, using $E^{-1}$ and conclude that $\text{null }T\subset\text{null }S$. Putting those together implies $\text{null }S=\text{null }T$. @@ -105,171 +105,68 @@ To do so, we will construct a linear map, $\tilde{E}\colon \text{range }T\to\text{range }S$, that is both injective and surjective (and thus invertible). - - - - -Suppose $\text{null } S = \text{null } T$. We'll try to construct the -required $E$. (That will show the “only if” part.) - -```mermaid -stateDiagram-v2 - direction LR - state "W" as s1 - state "W" as s2 - V --> s1 : S - V --> s2 : T - s2 --> s1 : E -``` - -What _is_ $E$? The diagram above isn't great but maybe gives some -idea. Starting from $V$, one can get to $W$ by using either $S$ or -$T$. (Note that the diagram “duplicates” $W$ -- that is, $W$ appears -twice.) The two maps $T$ and $S$ might give different answers: $E$ is -the map which “makes them the same.” $E$ takes the element of $W$ you -got by following $T$ and transforms it to the element of $W$ you got -by following $S$. - -Roughly speaking, the $E$ that we are looking for is “$`E=S\circ +Roughly speaking, the $\tilde{E}$ that we are looking for is “$`\tilde{E}=S\circ T^{-1}`$”. That is, given some element of $W$, we “take it back to $V$ -with $T^{-1}$ and then forwards again to $W$ using $S$.” Well, anyway, -that's the intuition -- but there are several questions to resolve. - -First, $T^{-1}$ might not exist if $T$ isn't surjective. We can't “go -back to $V$” from elements not in the range of $T$ because there's -nowhere for them to have come from. We will need to fix that by making -up somewhere for them to go when acted on by $E$. - -Second, $T^{-1}$ might not exist if $T$ isn't injective. That is, -it might be that there are _many_ elements of $V$ which are mapped by -$T$ to the same element of $W$. Then there is a choice of “where to go -back to.” Fortunately, it will turn out that this choice will not -matter. - -So let's start with what we can do. First of all, since we know what -to do with vectors in the range of $T$, let's start by separating that -out. Since $\text{range }T$ is a subspace of $W$, it is possible to -write $W$ as a direct sum $W = (\text{range }T)\oplus Z$ where $Z$ is -some other subspace of $W$. (This is Axler 2.33.) What this means is -that there's a way to write any vector in $W$ uniquely as the sum of a -vector in $\text{range }T$ and a vector in some other subspace $Z$ -which has no overlap with $\text{range }T$ (except the zero -vector). (The choice of $Z$ itself is not unique but we don't care -about that.) - -To define $E$ we need to say what it does to any element $w\in -W$. First, using the decomposition $W = \text{range }T\oplus Z$, write -that element as $w=u+z$ where $u\in\text{range }T$ and $z\in Z$. To -define the action of $E$ on $w$, we may separately define the action -of $E$ on $u$ and on $z$. - -Start with $u$. Since $u\in \text{range }T$, there must be some -$v\in V$ such that $T(v) = u$. Set - -```math -E(u) = S(v). -``` +with $T^{-1}$ and then forwards again to $W$ using $S$.” -We must be careful! There might be _another_ $v'\in V$ such that $T(v') -= u$. The rule above makes sense only if $S(v) = S(v')$ for any such -$v'$. Fortunately, that is true. Indeed, if $T(v')=T(v)$ then -$T(v'-v)=0$; in other words, $v'-v$ is in the null space of $T$. But -by supposition this is the null space of $S$. So $S(v'-v)=0$ and -therefore $S(v)=S(v')$. - -What about $z$? We will set +How do we make this more precise? Suppose we have some $w\in W$ that +is in the range of $T$. There must be _some_ $v\in V$ such that $T(v) += w$. We'd like to be able to define $\tilde{E}$ by ```math -E(z) = +\tilde{E}(w) = S(v). ``` -. - - - - -we know what $E$ has to do on vectors in the range of $T$ (it takes -$T(\vec{v})$ to $`S(\vec{v})`$) so we need to ensure (a) the this -action is linear and (b) that it does something sensible on other -vectors to make it invertible. Our usual approach to saying what we -mean by “other vectors” is to choose a basis for $W$ that contains a -basis for $\text{range }T$. - -That is: (1) choose a basis for the range of $T$, say $\vec{e}_1, -\dots, \vec{e}_m$, and then (2) extend to a basis for $W$, say -$`\vec{e}_1, \dots, \vec{e}_m, \vec{e}_{m+1}, \dots, \vec{e}_n`$. With -this basis, every element of $W$, say $\vec{w}$, can be written as -$`\vec{w} = \sum_i w_i \vec{e}_i`$, which is - -```math -\vec{w} = \sum_i w_i \vec{e}_i = \sum_{i=1}^{m} w_i \vec{e}_i + -\sum_{j=m+1}^n w_j \vec{e}_j -``` - -That is, every vector in $W$ can be written as the sum of an element -of $\text{range }T$ plus a vector not in the range of $T$. - -So now, for $\vec{w}\in W$, define $E$ in the following way. Write -$\vec{w} = \vec{u}+\vec{u}'$, where $\vec{u}\in\text{range }T$. Since -$\vec{u}\in\text{range }T$, there must be some place in $V$ from which -this came: there must be some $\vec{v}\in V$ (not necessarily unique) -such that $\vec{u} = T(\vec{v})$. We'd like to set $E(\vec{u}) = -S(\vec{v})$, but we have the problem that the $v$ is not necessarily -unique. Fortunately, the only ways that there can be multiple -$`\vec{v}`$s with $\vec{u}=T(\vec{v})$ is if they differ by elements -of the null space. - -That is, suppose $\vec{v}'$ is also such that $T(\vec{v}') = -\vec{u}$. Then $T(\vec{v}'-\vec{v}) = \mathbf{0}$; in other words, -$\vec{v}'-\vec{v}$ is in the null space of $T$. Therefore, by -supposition, $\vec{v}'-\vec{v}$ is in the null space of $S$. Therefore -$S(\vec{v}') = S(\vec{v})$ and so there is no ambiguity in setting -$E(\vec{u}) = S(\vec{v})$. - -We are nearly done with the definition of $E$. We want to define -$E(\vec{w})$ for some arbitrary $\vec{w}$, where however we have -written $\vec{w} = \vec{u}+\vec{u}'$, with $\vec{u}\in\text{range -T}$. We know what we want from $E(\vec{u})$ (it's wherever $S$ takes -the vector that $T$ sent to $\vec{u}$) but what about $E(\vec{u}')$? -We could just send that to zero but that would make $E$ -non-invertible. So instead let's leave it alone. In other words, $E$ -is defined by +The only problem is that there might be _another_ $v'\in V$ such that +$T(v') = w$. The rule above will only make sense if $S(v) = S(v')$ for +any such $v'$. Fortunately, that is true! Indeed, if $T(v')=T(v)$ then +$T(v'-v)=\mathbf{0}$; in other words, $v'-v$ is in the null space of +$T$. But by supposition this is the null space of $S$. So +$S(v'-v)=\mathbf{0}$ and therefore $S(v)=S(v')$. + +So we have constructed $\tilde{E}$ on any element of $\text{range +T}$. It is “go back to $V$ using $T$ and then forward to $W$ using +$S$, noting that it doesn't matter which element of $V$ you use.” + +To show that $\tilde{E}$ is an isomorphism, we need to show that it is +injective and surjective. It would fail to be injective if there were +$w,w'\in \text{range T}$, with $w\neq w'$, such that $\tilde{E}(w) = +\tilde{E}(w')$. But this would mean that $\tilde{E}(w-w')=\mathbf{0}$ +and therefore (by the definition of $\tilde{E}$) that +$\S(v'-v)=\mathbf{0}$ for some $v$ and $v'$ with $T(v)=w$ and +$T(v')=w'$. But _that_ means (since the null spaces are equal) that +$T(v'-v)=\mathbf{0}$ and therefore $w=w'$. So $\tilde{E}$ cannot fail +to be injective. + +It would fail to be injective if there were some $z\in \text{range }S$ +that is not the image of any $w$ under $\tilde{E}$. But it's easy to +find the required $w$: starting from $z$, “go back to $V$ using $S$ +and then forward to $W$ using $T$” following just the same argument above. + +We have shown that $\text{range }S$ and $text{range T}$ are +isomorphic. But to finish the argument, we must show that $\tilde{E}$ +has an extension to $W$. That is, we need to find an $E$ with +$E|_{\text{range }T}\tilde{E}$. + +Since $\text{range }T$ is a subspace, there is another subspace, say +$X_1$, such that $W=\text{range }T\oplus X$. (This is Axler 2.33, and +relies upon $W$ being finite-dimensional.) That is, one can find a +subspace $X$ such that any vector in $w$ can be written uniquely as +the sum of a vector in $\text{range }T$ and a vector in $X$. Likewise, +we can find a subspace $Y$ such that $W=\text{range }S\oplus Y$. + +Now -- and this is not very satisying -- since $W$ is +finite-dimensional, and since $\text{range }S\cong \text{range }T$, we +must have $\text{dim }X = \text{dim }Y$. Thus $X$ and $Y$ are +isomorphic. So there is some invertible $\tilde{F}\colon X\to Y$. + +Now we get to say what $E$ is. For $w\in W$, write $w = \tilde{w}+x$ +where $\tilde{w}\in\text{range }T$ and $x\in X$. Set ```math -E(\vec{w}) = S(\vec{v}) + \vec{u}', +E(w) = \tilde{E}(\tilde{w}) + \tilde{F}(x). ``` -where $\vec{w} = \vec{u}+\vec{u}'$ and $T(\vec{v}) = \vec{u}$. - -We _still_ have to show that $E$ is a linear map and invertible. To -show that it is linear is an annoying process of following through the -definition above for some $\alpha \vec{w}_1 + \beta \vec{w}_2$ (though -really it can't but be linear given that everything we have done is -linear). Why is it invertible? - -$E$ is invertible if it is injective and surjective. It can't be -non-injective on the bit of $W$ that is not the range of $T$, because -we've defined it to be the identity there. Suppose there were distinct -$\vec{u}, \vec{u}'$, both in the range of $T$ such that -$E(\vec{u})=E(\vec{u}')$. That would mean that there must be -$\vec{v},\vec{v}'\in V$, with $\vec{u}=T(\vec{v})$ and -$\vec{u}'=T(vec{v}')$, such that $S(\vec{v}) = S(\vec{v}')$. But that -means that $\vec{v}'-\vec{v}\in\text{null }S$, and hence -$\vec{v}'-\vec{v}\in\text{null }T$ and hence in fact -$T(\vec{v})=T(\vec{v}')$ or $\vec{u}=\vec{u}'$, in contradiction to -the supposition. So $E$ is injective. - -Is $E$ surjective? In other words, does every $\vec{w}\in\text{range -}S$ come from some $E(\vec{u})$? Yes: roughly, to find $\vec{u}$, go -back to $V$ (using $S$) and then forward to $W$ (using $T$): existence -and the required properties of $\vec{u}$ follow by running the same -argument above with $T$ and $S$ switched. - -Well, that's the "only if" direction. The "if" direction is -easy. Suppose $\vec{v}\in\text{null T}$, say. Then $S(\vec{v}) = -E(T(\vec{v})) = \mathbf{0}$, so $\vec{v}$ is in the null space of -$S$. The argument can be reversed, noting that $E$ is invertible. - ## Question 9 Suppose $V$ is finite-dimensional and $T\colon V\to W$ is a surjective