Merge pull request #34 from alan-turing-institute/2024-04-05

2024 04 05

exercises/axler-3d.md

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+# Exercises from Axler 3.C
+
+1, 2, 6, 9, 18, 21
+
+## Question 1
+
+Suppose $T\in\mathcal{L}(V, w)$ is invertible. Show that $T^{-1}$ is
+invertible and $(T^{-1})^{-1} = T$.
+
+### Answer
+
+When asked to show that something has a certain property, it's often
+best to go back to the definition of that property. A linear map
+$R:V\to W$ is invertible (this is definition 3.59) if there exists an
+$S:W\to V$ such that $RS$ is the identity on $W$ and $SR$ is the
+identity on $V$.
+
+So, in order to show that some linear map (say $T^{-1}$) is
+invertible, we must exhibit another linear map with the required
+properties.
+
+The obvious candidate for an inverse of $T^{-1}$ is $T$. Let's check
+whether it is. We need to check that $T^{-1}T$ is the identity on $V$
+and that $TT^{-1}$ is the identity on $W$. And indeed they are --
+because _those_ conditions are precisely the meaning of $T^{-1}$
+(which we are assured exists by the question).
+
+## Question 2
+
+Suppose $T\in \mathcal{L}(U, V)$ and $S\in \mathcal{L}(V, W)$ are both
+invertible. Prove that $ST$ is invertible and that $(ST)^{-1} =
+T^{-1}S^{-1}$.
+
+### Answer
+
+To prove that something is invertible we must exhibit its
+inverse. Fortunately, the question has told us what the inverse is, so
+what we need to do is check that it does in fact satisfy the
+conditions for being an inverse.
+
+That is, we need to show: (1) that $\bigl(ST\bigr) \bigl(
+T^{-1}S^{-1}\bigr)$ is the identity on $W$; and (2) that $\bigl(
+T^{-1}S^{-1}\bigr) \bigl(ST\bigr)$ is the identity on $U$.
+
+To show (1), note that composition of linear maps is associative --
+that is, it doesn't matter where we put the parentheses. Thus, roughly
+speaking, we can “cancel the inner $`T`$s” and then, having got rid of
+those, “cancel the outer $`S`$s.”
+
+More carefully, $\bigl(ST\bigr) \bigl( T^{-1}S^{-1}\bigr) = S(T
+T^{-1})S^{-1}$ because composition is associative. And that is
+$S\mathbf{1}_V S^{-1}$, which is $SS^{-1}$ which is
+$\mathbf{1}_W$. The same argument works for (2).
+
+## Question 6
+
+Suppose that $W$ is finite-dimensional and $S, T\in\mathcal{L}(V,
+W)$. Prove that $\text{null } S = \text{null } T$ if and only if there
+exists an invertible $E\in\mathcal{L}(W)$ such that $S=ET$.
+
+### Answer
+
+Suppose $\text{null } S = \text{null } T$. We'll try to construct the
+required $E$. Roughly speaking, 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
+
+```math
+E(\vec{w}) = S(\vec{v}) + \vec{u}',
+```
+
+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
+linear map of $V$ onto $W$. Prove that there is a subspace $U$ of $V$
+such that $T\rvert_U$ is an isomorphism of $U$ onto $W$.
+
+### Answer
+
+Question 6 was about splitting up $W$ into “the range of $T$ plus
+another bit.” This question is sort of the complement: we're splitting
+$V$ into “the null space of $T$ plus another bit.”
+
+So, let $(\vec{e}_1, \dots, \vec{e}_m, \vec{f}_1, \dots, \vec{f}_n)$ be
+a basis of $V$ such that $(\vec{e}_1, \dots, \vec{e}_m)$ is a basis for
+$\text{null }T$ and the $`\vec{f}`$s extend that to a basis of the
+full space.
+
+I claim that a subspace satisfying the conditions of the question is
+$U = \text{span }(\vec{f}_1, \dots, \vec{f}_n)$. First of all, it _is_
+a subspace (since it’s the span of some vectors). The map $T|_U$ is
+also surjective. For, given some $\vec{w}\in W$, then, since the
+original $T$ was surjective, there must be $\vec{v}\in V$ with
+$T(\vec{v}) = \vec{w}$. Now write $\vec{v}=\vec{u}+\vec{\nu}$ where
+the $\vec{u}$ is a combination of the $`\vec{f}`$s and the $\vec{\nu}$
+is a combination of the $`\vec{e}`$s (ie, in the null space). Since
+$T(\vec{\nu})=\mathbf{0}$ we must have $\vec{w} = T(\vec{v}) =
+T(\vec{u}) = T|_U(\vec{u}$. And since $\vec{u}\in U$, we have shown
+that $T|_U$ is surjective.
+
+Then, suppose $\vec{v}_1,\vec{v}_2\in U$ are distinct; it follows that
+$T|_U(\vec{v}_1) \neq T|_U(\vec{v}_2)$, otherwise their difference
+would be in $\text{null }T$, and hence that $T|_U$ is injective.
+
+Finally, since $T|_U$ is surjective and injective, it is an
+isomorphism.
+
+## Question 18
+
+Show that $V$ and $\mathcal{L}(\mathbf{F}, V)$ are isomorphic vector
+spaces.
+
+### Answer
+
+We must exhibit an invertible linear map between $V$ and
+$\mathcal{L}(\mathbf{F}, V)$. 
+
+What is $\mathcal{L}(\mathbf{F}, V)$? It is the space of linear maps
+from $\mathbf{F}$ to $V$. That is, an element of
+$\mathcal{L}(\mathbf{F}, V)$ is a rule which, given a number in
+$\mathbf{F}$, produces a vector in $V$. Suppose
+$\tilde{w}\in\mathcal{L}(\mathbf{F}, V)$ is one such rule. Then, for
+any $\alpha\in\mathbf{F}$, $`\tilde{w}(\alpha) =
+\tilde{w}(\alpha\times 1) = \alpha\tilde{w}(1)`$. In other words, any
+$\tilde{w}\in\mathcal{L}(\mathbf{F}, V)$ is defined entirely by its
+action on $1$. Indeed, $\tilde{w}(1)$ is an element of $V$, so we can
+say that any element of $\mathcal{L}(\mathbf{F}, V)$ is defined by a
+specific vector in $V$. That's the informal story.
+
+More carefully, consider the map, $\Omega\colon
+V\to\mathcal{L}(\mathbf{F}, V)$ which acts in the following way: for
+any $\vec{v}\in V$, set $\Omega(\vec{v})$ to be the map
+
+```math
+\begin{aligned}
+\Omega(\vec{v}) : \mathbf{F} &\to  V \\
+\alpha &\mapsto \alpha \vec{v}.
+\end{aligned}
+```
+
+That is to say, $\Omega(\vec{v})$ is an element of
+$\mathcal{L}(\mathbf{F}, V)$, hence a map; specifically the map which
+takes any $\alpha\in\mathbf{F}$ to the vector $\alpha\vec{v}$.
+
+I claim the map $\Omega$ is both injective and surjective. Suppose
+$\vec{v}, \vec{v}'\in V$ are distinct. Then clearly
+$\Omega(\vec{v}) \neq \Omega(\vec{v}')$, since they take $1$ to
+different vectors. That shows injectivity of $\Omega$.
+
+Suppose $\tilde{w}\in \mathcal{L}(\mathbf{F}, V)$ is some map. Then
+$\tilde{w}(1)$ is some element of $V$, and indeed $\Omega(\tilde{w}(1)) =
+\tilde{w}$. That shows surjectivity of $\Omega$.
+
+## Question 21
+
+Suppose $n$ is a positive integer and $A_{jk}\in\mathbf{F}$ for
+$j,k\in\{1,\dots,n\}$. Prove that the following are equivalent:
+
+(a) The only solution to the homogeneous system of equations
+
+```math
+\sum_k A_{ik} x_k = 0 \quad (i\in\{1,\dots, n\})
+```
+
+is the trivial solution $x_1 = \dotsb = x_n = 0$.
+
+(b) For every $c_1,\dots,c_n\in\mathbf{F}$, there exists a solution to
+the system of equations
+
+```math
+\sum_k A_{ik} x_k = c_i \quad (i\in\{1,\dots, n\}).
+```
+
+### Answer
+
+Consider the vector space $\mathbf{F}^n$ and let $A$ be the linear map
+in $\mathcal{L}(\mathbf{F}^n, \mathbf{F}^n)$ whose matrix
+representation in the canonical basis of $\mathbf{F}^n$ is
+$A_{jk}$.
+
+Then (a) is equivalent to the statement that $A$ is injective; and $b$
+is equvalent to the statement that $A$ is surjective. But we know
+(Axler 3.65) that, for linear maps between finite-dimensional vector
+spaces of the same dimension, each of these properties implies the
+other.