From 3c0400ab2509e71c05ad25a7cf689aa1814ce58c Mon Sep 17 00:00:00 2001 From: James Geddes Date: Thu, 28 Mar 2024 09:39:08 +0000 Subject: [PATCH 01/15] Add start of solutions to 3D --- exercises/axler-3d.md | 53 ++++++++++++++++++++++++++++++++++++++++ implementation/README.md | 19 +++++++++----- 2 files changed, 66 insertions(+), 6 deletions(-) create mode 100644 exercises/axler-3d.md diff --git a/exercises/axler-3d.md b/exercises/axler-3d.md new file mode 100644 index 0000000..55b9fd2 --- /dev/null +++ b/exercises/axler-3d.md @@ -0,0 +1,53 @@ +# Exercises from Axler 3.C + +1, 2, 6, 9, 18, 20 + +## 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 inverse on $V$ +and that $TT^{-1}$ is the inverse 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). diff --git a/implementation/README.md b/implementation/README.md index d4633bd..efe18bc 100644 --- a/implementation/README.md +++ b/implementation/README.md @@ -101,10 +101,17 @@ Using Numpy. ### Racket -Naive implementation, boxed floats. +1. `naive`: Boxed floats and generic numeric operations.. +2. `naive-cons`: Replacing `struct` with a pair. +3. `flo`: Using floating-point * and + in the inner loop of matrix + multiply and flvectors. (But still safe.) + +All on Apple M2. + +| Device | s | ops/s | +|:-------------|-----:|----------:| +| `naive` | 39.4 | 426,272 | +| `naive-cons` | 26.9 | 623,688 | +| `flo` | 11.4 | 1,472,000 | +| `unsafe` | 5.23 | 3,208,000 | -| Device | s | ops/s | -|:-------------|------:|--------:| -| Intel i7 | | | -| Apple M1 Pro | | | -| Apple M2 | 39.36 | 426,272 | From 8a93ead6917cd69a5ff0343c2ee9f4380c2a1f7c Mon Sep 17 00:00:00 2001 From: James Geddes Date: Fri, 29 Mar 2024 08:27:36 +0000 Subject: [PATCH 02/15] Fix typos --- exercises/axler-3d.md | 16 ++++++++-------- 1 file changed, 8 insertions(+), 8 deletions(-) diff --git a/exercises/axler-3d.md b/exercises/axler-3d.md index 55b9fd2..4f7afba 100644 --- a/exercises/axler-3d.md +++ b/exercises/axler-3d.md @@ -1,6 +1,6 @@ # Exercises from Axler 3.C -1, 2, 6, 9, 18, 20 +1, 2, 6, 9, 18, 21 ## Question 1 @@ -20,8 +20,8 @@ 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 inverse on $V$ -and that $TT^{-1}$ is the inverse on $W$. And indeed they are -- +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). @@ -38,16 +38,16 @@ 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 +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.” +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 +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). From e8695ddfaf2a89f6f20ce7a3707b79d06e39b2e0 Mon Sep 17 00:00:00 2001 From: James Geddes Date: Fri, 29 Mar 2024 08:49:29 +0000 Subject: [PATCH 03/15] Working on 6. --- exercises/axler-3d.md | 24 ++++++++++++++++++++++++ 1 file changed, 24 insertions(+) diff --git a/exercises/axler-3d.md b/exercises/axler-3d.md index 4f7afba..521f4a4 100644 --- a/exercises/axler-3d.md +++ b/exercises/axler-3d.md @@ -51,3 +51,27 @@ 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 construction is: (1) choose a basis for $\text{range } T$, +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 $\sum_i w_i \vec{e}_i$, which is $\sum_{i=1}^{m} w_i +\vec{e}_i + \sum_{j=m+1}^n w_j \vec{e}_j$ -- that is, the sum of an +element of $\text{range }T$ plus a vector not in the range of $T$. + + From c9d70d43a79704ab59da151972bc3f4349365cd1 Mon Sep 17 00:00:00 2001 From: James Geddes Date: Fri, 29 Mar 2024 08:55:04 +0000 Subject: [PATCH 04/15] Math typos --- exercises/axler-3d.md | 14 +++++++------- 1 file changed, 7 insertions(+), 7 deletions(-) diff --git a/exercises/axler-3d.md b/exercises/axler-3d.md index 521f4a4..aba7780 100644 --- a/exercises/axler-3d.md +++ b/exercises/axler-3d.md @@ -62,16 +62,16 @@ exists an invertible $E\in\mathcal{L}(W)$ such that $S=ET$. 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 +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 construction is: (1) choose a basis for $\text{range } T$, -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 $\sum_i w_i \vec{e}_i$, which is $\sum_{i=1}^{m} w_i -\vec{e}_i + \sum_{j=m+1}^n w_j \vec{e}_j$ -- that is, the sum of an +Our usual construction 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 $`\sum_{i=1}^{m} w_i +\vec{e}_i + \sum_{j=m+1}^n w_j \vec{e}_j`$ -- that is, the sum of an element of $\text{range }T$ plus a vector not in the range of $T$. From d1439f5c8be9d0109d88b7feffd27a6e6dcd8bdd Mon Sep 17 00:00:00 2001 From: James Geddes Date: Fri, 29 Mar 2024 08:57:18 +0000 Subject: [PATCH 05/15] Typos --- exercises/axler-3d.md | 10 +++++++--- 1 file changed, 7 insertions(+), 3 deletions(-) diff --git a/exercises/axler-3d.md b/exercises/axler-3d.md index aba7780..4479cfe 100644 --- a/exercises/axler-3d.md +++ b/exercises/axler-3d.md @@ -70,8 +70,12 @@ Our usual construction 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 $`\sum_{i=1}^{m} w_i -\vec{e}_i + \sum_{j=m+1}^n w_j \vec{e}_j`$ -- that is, the sum of an -element of $\text{range }T$ plus a vector not in the range of $T$. +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 $T$. From 7005df3293e0c034e7496e9d64167073d93fc79e Mon Sep 17 00:00:00 2001 From: James Geddes Date: Fri, 29 Mar 2024 09:15:48 +0000 Subject: [PATCH 06/15] Lots of maths. --- exercises/axler-3d.md | 37 +++++++++++++++++++++++++++++++------ 1 file changed, 31 insertions(+), 6 deletions(-) diff --git a/exercises/axler-3d.md b/exercises/axler-3d.md index 4479cfe..5963aed 100644 --- a/exercises/axler-3d.md +++ b/exercises/axler-3d.md @@ -64,13 +64,15 @@ 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. +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 $\text{range }T$. -Our usual construction 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 +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 + @@ -79,3 +81,26 @@ can be written as $`\vec{w} = \sum_i w_i \vec{e}_i`$, which is 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 $E(\vec{u}')$? From 66863427a483d479eb9b65c3f53861f5f76b82c4 Mon Sep 17 00:00:00 2001 From: James Geddes Date: Fri, 29 Mar 2024 09:54:25 +0000 Subject: [PATCH 07/15] Question 6 is quite long. --- exercises/axler-3d.md | 44 ++++++++++++++++++++++++++++++++++++++++--- 1 file changed, 41 insertions(+), 3 deletions(-) diff --git a/exercises/axler-3d.md b/exercises/axler-3d.md index 5963aed..49fb206 100644 --- a/exercises/axler-3d.md +++ b/exercises/axler-3d.md @@ -93,7 +93,7 @@ $`\vec{v}`$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}$; in other words, +\vec{u}$. 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 @@ -101,6 +101,44 @@ $E(\vec{u}) = S(\vec{v})$. 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 +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 $E(\vec{u}')$? +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. From b9a5f65fdf0b47ad734aaff985211320ea4625ef Mon Sep 17 00:00:00 2001 From: James Geddes Date: Fri, 29 Mar 2024 10:04:25 +0000 Subject: [PATCH 08/15] Start question 9. --- exercises/axler-3d.md | 20 ++++++++++++++++++++ 1 file changed, 20 insertions(+) diff --git a/exercises/axler-3d.md b/exercises/axler-3d.md index 49fb206..60b4340 100644 --- a/exercises/axler-3d.md +++ b/exercises/axler-3d.md @@ -142,3 +142,23 @@ 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 +$T|_U = \text{span }(\vec{f}_1, \dots, \vec{f}_n$. From d77f510c99fe7ea513ddb4cc45a86775f70371c2 Mon Sep 17 00:00:00 2001 From: James Geddes Date: Fri, 29 Mar 2024 16:02:08 +0000 Subject: [PATCH 09/15] Start question 9. --- exercises/axler-3d.md | 47 +++++++++++++++++++++++++++++++++++++++++-- 1 file changed, 45 insertions(+), 2 deletions(-) diff --git a/exercises/axler-3d.md b/exercises/axler-3d.md index 60b4340..3914d52 100644 --- a/exercises/axler-3d.md +++ b/exercises/axler-3d.md @@ -156,9 +156,52 @@ 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 +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 -$T|_U = \text{span }(\vec{f}_1, \dots, \vec{f}_n$. +$U = \text{span }(\vec{f}_1, \dots, \vec{f}_n)$. 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 $\vec{v}=\vec{u}+\vec{u}'$ where the +$\vec{u}$ is a combination of the $`\vec{e}`$s and the $\vec{u}'$ is a +combination of the $`\vec{f}`$s. Since $T(\vec{u})=\mathbf{0}$ we must +have $T(\vec{v}) = T(\vec{u}')$. 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. + +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 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 $1$. + +Consider the map, $\Omega\colon V\mathcal{L}(\mathbf{F}, V)$ 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}) \colon \mathbf{F} &\to V \\ +\Omega(\vec{v}) \colon \alpha \mapsto \alpha \vec{v}. +\end{algined} +``` From 198a96b0f97d01838064e954efd6077d6089fac9 Mon Sep 17 00:00:00 2001 From: James Geddes Date: Fri, 29 Mar 2024 16:07:52 +0000 Subject: [PATCH 10/15] Typos --- exercises/axler-3d.md | 22 ++++++++++++---------- 1 file changed, 12 insertions(+), 10 deletions(-) diff --git a/exercises/axler-3d.md b/exercises/axler-3d.md index 3914d52..2a8ae52 100644 --- a/exercises/axler-3d.md +++ b/exercises/axler-3d.md @@ -165,17 +165,19 @@ $U = \text{span }(\vec{f}_1, \dots, \vec{f}_n)$. 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 $\vec{v}=\vec{u}+\vec{u}'$ where the -$\vec{u}$ is a combination of the $`\vec{e}`$s and the $\vec{u}'$ is a -combination of the $`\vec{f}`$s. Since $T(\vec{u})=\mathbf{0}$ we must -have $T(\vec{v}) = T(\vec{u}')$. And since $\vec{u}'\in U$, we have -shown that $T|_U$ is surjective. +$T(\vec{v}) = \vec{w}$. 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. -since $T|_U$ is surjective and injective, it is an isomorphism. +Finally, since $T|_U$ is surjective and injective, it is an +isomorphism. ## Question 18 @@ -190,12 +192,12 @@ $\mathcal{L}(\mathbf{F}, V)$. What is $\mathcal{L}(\mathbf{F}, V)$? It 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 +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 $1$. -Consider the map, $\Omega\colon V\mathcal{L}(\mathbf{F}, V)$ which +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 @@ -203,5 +205,5 @@ $\Omega(\vec{v})$ to be the map \begin{aligned} \Omega(\vec{v}) \colon \mathbf{F} &\to V \\ \Omega(\vec{v}) \colon \alpha \mapsto \alpha \vec{v}. -\end{algined} +\end{aligned} ``` From aaa0ab2ab25a2db521f402b2aa772079ba63af9e Mon Sep 17 00:00:00 2001 From: James Geddes Date: Fri, 29 Mar 2024 16:10:26 +0000 Subject: [PATCH 11/15] Typos --- exercises/axler-3d.md | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/exercises/axler-3d.md b/exercises/axler-3d.md index 2a8ae52..9bc3f15 100644 --- a/exercises/axler-3d.md +++ b/exercises/axler-3d.md @@ -168,7 +168,7 @@ 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{\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. @@ -203,7 +203,7 @@ $\Omega(\vec{v})$ to be the map ```math \begin{aligned} -\Omega(\vec{v}) \colon \mathbf{F} &\to V \\ -\Omega(\vec{v}) \colon \alpha \mapsto \alpha \vec{v}. +\Omega(\vec{v}) : \mathbf{F} &\to V \\ +\alpha &\mapsto \alpha \vec{v}. \end{aligned} ``` From 5380e6102249e2bd21e85e2c018e87738d82bb85 Mon Sep 17 00:00:00 2001 From: James Geddes Date: Fri, 29 Mar 2024 16:54:42 +0000 Subject: [PATCH 12/15] Question 21 --- exercises/axler-3d.md | 52 ++++++++++++++++++++++++++++++++++++++----- 1 file changed, 46 insertions(+), 6 deletions(-) diff --git a/exercises/axler-3d.md b/exercises/axler-3d.md index 9bc3f15..5a2ab36 100644 --- a/exercises/axler-3d.md +++ b/exercises/axler-3d.md @@ -189,17 +189,21 @@ spaces. We must exhibit an invertible linear map between $V$ and $\mathcal{L}(\mathbf{F}, V)$. -What is $\mathcal{L}(\mathbf{F}, V)$? It is a rule which, given a -number in $\mathbf{F}$, produces a vector in $V$. Suppose +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 $1$. +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. -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 +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} @@ -207,3 +211,39 @@ $\Omega(\vec{v})$ to be the map \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. + +Suppose $\tilde{w}\in \mathcal{L}(\mathbf{F}, V)$ is some map. Then +$\tilde{w}(1)$ is some element of $V$, and indeed $\Omega(\tilde{1}) = +\omega$. That shows surhectivity. + +## 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} = 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} = c_i \quad (i\in\{1,\dots, n\}). +``` + + +### Answer From 7c7378f0db4d6d71a08d0675632b197af7270555 Mon Sep 17 00:00:00 2001 From: James Geddes Date: Fri, 29 Mar 2024 16:57:23 +0000 Subject: [PATCH 13/15] Typos --- exercises/axler-3d.md | 10 +++++----- 1 file changed, 5 insertions(+), 5 deletions(-) diff --git a/exercises/axler-3d.md b/exercises/axler-3d.md index 5a2ab36..aa3e921 100644 --- a/exercises/axler-3d.md +++ b/exercises/axler-3d.md @@ -219,11 +219,11 @@ 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. +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 $\Omega(\tilde{1}) = -\omega$. That shows surhectivity. +$\tilde{w}(1)$ is some element of $V$, and indeed $\Omega(\tilde{w}(1)) = +\tilde{w}$. That shows surjectivity of $\Omega$. ## Question 21 @@ -233,7 +233,7 @@ $j,k\in\{1,\dots,n\}$. Prove that the following are equivalent: (a) The only solution to the homogeneous system of equations ```math -\sum_k A_{ik} = 0 \quad (i\in\{1,\dots, n\}) +\sum_k A_{ik} x_k = 0 \quad (i\in\{1,\dots, n\}) ``` is the trivial solution $x_1 = \dotsb = x_n = 0$. @@ -242,7 +242,7 @@ $j,k\in\{1,\dots,n\}$. Prove that the following are equivalent: the system of equations ```math -\sum_k A_{ik} = c_i \quad (i\in\{1,\dots, n\}). +\sum_k A_{ik} x_k = c_i \quad (i\in\{1,\dots, n\}). ``` From cb4facd895d485c2d97dd9f917b7961d3cf779e1 Mon Sep 17 00:00:00 2001 From: James Geddes Date: Fri, 29 Mar 2024 17:19:01 +0000 Subject: [PATCH 14/15] Solution to 21 --- exercises/axler-3d.md | 14 ++++++++++++-- 1 file changed, 12 insertions(+), 2 deletions(-) diff --git a/exercises/axler-3d.md b/exercises/axler-3d.md index aa3e921..9e1976d 100644 --- a/exercises/axler-3d.md +++ b/exercises/axler-3d.md @@ -236,7 +236,7 @@ $j,k\in\{1,\dots,n\}$. Prove that the following are equivalent: \sum_k A_{ik} x_k = 0 \quad (i\in\{1,\dots, n\}) ``` - is the trivial solution $x_1 = \dotsb = x_n = 0$. +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 @@ -245,5 +245,15 @@ the system of equations \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. From 2de99804dd6da3efce31b5a5643d1a8135fc3abe Mon Sep 17 00:00:00 2001 From: James Geddes Date: Fri, 29 Mar 2024 17:24:19 +0000 Subject: [PATCH 15/15] Final version of exercises --- exercises/axler-3d.md | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/exercises/axler-3d.md b/exercises/axler-3d.md index 9e1976d..921b728 100644 --- a/exercises/axler-3d.md +++ b/exercises/axler-3d.md @@ -203,7 +203,7 @@ 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 +any $\vec{v}\in V$, set $\Omega(\vec{v})$ to be the map ```math \begin{aligned} @@ -213,7 +213,7 @@ any $\vec{v}\in V$ set $\Omega(\vec{v})$ to be the map ``` That is to say, $\Omega(\vec{v})$ is an element of -$\mathcal{L}(\mathbf{F}, V)$, hence a map, specifically the map which +$\mathcal{L}(\mathbf{F}, V)$, 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 @@ -252,7 +252,7 @@ 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 $b$ +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