Tensor Algebras - Cooperstein Theorem 10.8

[Math Amateur]

I am reading Bruce N. Coopersteins book: Advanced Linear Algebra (Second Edition) ... ...

I am focused on Section 10.3 The Tensor Algebra ... ...

I need help in order to get a basic understanding of Theorem 10.8 which is a Theorem concerning the direct sum of a family of subspaces as a solution to a UMP ... the theorem is preliminary to tensor algebras ...

I am struggling to understand how the function $G$ as defined in the proof actually gives us $G \circ \epsilon_i = g_i$ ... ... if I see the explicit mechanics of this I may understand the functions involved better ... and hence the whole theorem better ...


Theorem 10.8 (plus some necessary definitions and explanations) reads as follows:

In the above we read the following:


" ... ... Then define


$G(f) = \sum_{j = 1}^t g_{i_j} (f(i_j))$


We leave it to the reader to show that this is a linear transformation and if $G$ exists then it must be defined in this way, that is, it is unique. ... ... "


Can someone please help me to ...

(1) demonstrate explicitly, clearly and in detail that $G(f) = \sum_{j = 1}^t g_{i_j} (f(i_j))$ satisfies $G \circ \epsilon_i = g_i$ (if I understand the detail of this then I may well understand the functions involved better, and in turn, understand the theorem better ...)


(2) show that $G$ is a linear transformation and, further, that if $G$ exists then it must be defined in this way, that is, it is unique.



Hope that someone can help ...

Peter

 

[andrewkirk]

We are given that $spt(f)=\{i_1,...,i_t\}$. That means that $\exists v_{i_1}\in V_{i_1},...,v_{i_t}\in V_{i_t}$ such that
$$f=\epsilon_{i_1}(v_{i_1})+...+\epsilon_{i_t}(v_{i_t})$$

Then for $k\in \{1,...,t\}$ we have, by replacing $f$ by $\epsilon_{i}(v_{i})$ in the definition of $G(f)$:

$$G\circ\epsilon_{i}(v_{i})\equiv G(\epsilon_{i}(v_{i}))\equiv\sum_{j=1}^t g_{i_j}(\epsilon_{i}(v_{i})(i_j))$$

In your earlier thread $\epsilon_k$ was defined as the function from $V_k$ to the direct product $V$ that maps $v_k$ to
$(0,0,...0,v_k,0,...,0)$ where $v_k$ is in the $k$th position. That only covers finite direct sums. However it looks from the above as though Cooperstein was - between the two sections you quoted - moved on to defining and allowing infinite direct sums (because of his use of an index set $I$ of unspecified cardinality, rather than just labelling the component spaces as $V_1$ to $V_n$). That means that $\epsilon_k$ needs an appropriately modified definition. I'm guessing the definition he's using is something like that $\epsilon_i:V_i\to V$ such that $\epsilon_i(v_i)(j)$ is zero for all $j\in I$ except $i$, for which it gives $v_i$.

Applying that to the above equation, we have

$$G\circ\epsilon_{i}(v_{i})\equiv \sum_{j=1}^t g_{i_j}(\epsilon_{i}(v_{i})(i_j))
=g_{i}(v_i)$$
as required.

 

[Math Amateur]

Thanks so much, Andrew ...

Just working through your post and reflecting on what you have said ...

Thanks again for your help ...

Peter

 

[andrewkirk]

And the linearity of $G$ follows from the linearity of the $g_k$. Say we have $f_1=\sum_{k=1}^{t_1}\epsilon_{1k}(v_{1k}),f_2=\sum_{k=1}^{t_2}\epsilon_{2k}(v_{2k})$ where $spt(f_1)=a_{11},...,a_{1t_1},spt(f_2)=a_{21},...,a_{2t_2}$
so that $spt(f_1+f_2)\subseteq spt(f_1)\cup spt(f_2)=i_1...i_t$ ($t\leq t_1+t_2$) so that we can write
$$f_1=\sum_{k=1}^{t}\epsilon_{3k}(u_{1k}),f_2=\sum_{k=1}^{t}\epsilon_{3k}(u_{2k})$$
where, for $r\in\{1,2\}$, $u_{rk}=v_{rj}$ for the $j$ such that $a_{rj}=i_k$ if $i_k\in spt(f_r)$ and otherwise $u_{rk}=0$.

Then we have
$$G(f_1+f_2)=
G\left(\sum_{k=1}^{t}\epsilon_{3k}(u_{1k})+\sum_{k=1}^{t}\epsilon_{3k}(u_{2k})\right)
=\sum_{j=1}^t g_{i_j}\left(\sum_{k=1}^{t}\epsilon_{3k}(u_{1k})(i_j)+\sum_{k=1}^{t}\epsilon_{3k}(u_{2k})(i_j)\right)$$

$$=\sum_{j=1}^t g_{i_j}\left(\sum_{k=1}^{t}\epsilon_{3k}(u_{1k})(i_j)\right)+
\sum_{j=1}^t g_{i_j}\left(\sum_{k=1}^{t}\epsilon_{3k}(u_{2k})(i_j)\right)
$$

$$=\sum_{j=1}^t g_{i_j}\left(f_1(i_j)\right)+
\sum_{j=1}^t g_{i_j}\left(f_2(i_j)\right)
=G(f_1)+G(f_2)$$

Proving that $G(\lambda f)=\lambda G(f)$ follows the general pattern of this proof but is much easier.

 

[andrewkirk]

Finally, uniqueness. We use the fact that $\{\epsilon_i(v_i)\ :\ i\in I\wedge v_i\in V_i\}$ is a spanning set for $V$. I used that in the first line of the previous post but omitted to point that out. The proof is straightforward, based on the fact that all elements of $V$ have finite support.

Say we have another linear map $G':V\to W$ such that $\forall i\in I:\ G'\circ \epsilon_i =g_i$. Then, for any $f\in V$, writing $f=\sum_{k=1}^t \epsilon_k(v_k)$, we have

$$G'(f)=G'\left(\sum_{k=1}^t \epsilon_k(v_k)\right)
=\sum_{k=1}^t G'\circ \epsilon_k(v_k)
=\sum_{k=1}^t g_k(v_k)
=\sum_{k=1}^t G\circ \epsilon_k(v_k)
=G\left(\sum_{k=1}^t \epsilon_k(v_k)\right)=G(f)
$$

So $G'=G$.

 

[Math Amateur]

We are given that $spt(f)=\{i_1,...,i_t\}$. That means that $\exists v_{i_1}\in V_{i_1},...,v_{i_t}\in V_{i_t}$ such that
$$f=\epsilon_{i_1}(v_{i_1})+...+\epsilon_{i_t}(v_{i_t})$$

Then for $k\in \{1,...,t\}$ we have, by replacing $f$ by $\epsilon_{i}(v_{i})$ in the definition of $G(f)$:

$$G\circ\epsilon_{i}(v_{i})\equiv G(\epsilon_{i}(v_{i}))\equiv\sum_{j=1}^t g_{i_j}(\epsilon_{i}(v_{i})(i_j))$$

In your earlier thread $\epsilon_k$ was defined as the function from $V_k$ to the direct product $V$ that maps $v_k$ to
$(0,0,...0,v_k,0,...,0)$ where $v_k$ is in the $k$th position. That only covers finite direct sums. However it looks from the above as though Cooperstein was - between the two sections you quoted - moved on to defining and allowing infinite direct sums (because of his use of an index set $I$ of unspecified cardinality, rather than just labelling the component spaces as $V_1$ to $V_n$). That means that $\epsilon_k$ needs an appropriately modified definition. I'm guessing the definition he's using is something like that $\epsilon_i:V_i\to V$ such that $\epsilon_i(v_i)(j)$ is zero for all $j\in I$ except $i$, for which it gives $v_i$.

Applying that to the above equation, we have

$$G\circ\epsilon_{i}(v_{i})\equiv \sum_{j=1}^t g_{i_j}(\epsilon_{i}(v_{i})(i_j))
=g_{i}(v_i)$$
as required.

Hi Andrew,

Thanks again for your help ...

Just a couple of clarifications, though ...

1. I know that $f$ has finite support which means $f$ is non-zero at only finite $i \in I$ ... but ... I cannot follow how/why we have

$$f=\epsilon_{i_1}(v_{i_1})+...+\epsilon_{i_t}(v_{i_t})$$


Could you please explain ( perhaps, if you would, slowly and in detail ,,, ) why/how this is true ... and maybe what it means ..



2. You write:

" ... ... something like that $\epsilon_i:V_i\to V$ such that $\epsilon_i(v_i)(j)$ is zero for all $j\in I$ except $i$, for which it gives $v_i$. ... ... "


$\epsilon_i$ has domain $V_i$ and so I understand an expression like $\epsilon_i (v_i)$ ... but your expression $\epsilon_i(v_i)(j)$ has two arguments, namely $v_i$ and $j$ ... ? ... can you explain what is going on ...


Sorry to be slow and perhaps over-careful ... but I am trying to ensure that I fully understand the material ...

Peter

 

[andrewkirk]

1. I know that $f$ has finite support which means $f$ is non-zero at only finite $i \in I$ ... but ... I cannot follow how/why we have

$$f=\epsilon_{i_1}(v_{i_1})+...+\epsilon_{i_t}(v_{i_t})$$


Could you please explain ( perhaps, if you would, slowly and in detail ,,, ) why/how this is true ... and maybe what it means ..

It comes from the adaptation of the projection functions $\pi_i:V\to V_i$, defined in your earlier thread, to the infinite-dimensional case. A little reflection shows that the natural adaptation (which may perhaps be in the intervening passages of Cooperstein) is to define $\pi_i$ by $\pi_i(f)\equiv f(i)$.

There is a little work to be done to re-prove (a) and (b) from your External Direct Sum thread for the infinite-dimensional case (although I note that Cooperstein didn't even bother proving them in the finite-dimensional case. I think he's a bit slack.), but it should be pretty straightforward.

Taking that as read, we proceed as follows:

$f:I\to \bigcup_{i\in I}V_i$ has finite support, so let the support be $i_1,...,i_t$ and let $v_{i_k}\equiv f(i_k)$.

Next, we use (b)

$$\sum_{i\in I}\epsilon_i\circ \pi_i=I_V$$

to get

$$f=I_Vf=\sum_{i\in I}\epsilon_i\circ \pi_i(f)
=\sum_{i\in I}\epsilon_i\left(\pi_i(f)\right)
=\sum_{i\in I}\epsilon_i\left(f(i)\right)$$
Note that, by the linearity of $\epsilon_i$, the elements of this last sum are all zero except when $i\in\{i_1,...,i_t\}$, so we have

$$f=\sum_{k=1}^t\epsilon_{i_k}\left(f(i_k)\right)
=\sum_{k=1}^t\epsilon_{i_k}\left(v_{i_k}\right)$$

as required.

2. You write:

" ... ... something like that $\epsilon_i:V_i\to V$ such that $\epsilon_i(v_i)(j)$ is zero for all $j\in I$ except $i$, for which it gives $v_i$. ... ... "

$\epsilon_i$ has domain $V_i$ and so I understand an expression like $\epsilon_i (v_i)$ ... but your expression $\epsilon_i(v_i)(j)$ has two arguments, namely $v_i$ and $j$ ... ? ... can you explain what is going on ...

$\epsilon_i(v_i)$ is an element of the direct sum $V$. Recall that the elements of the direct sum are functions from the index set $I$ to $\bigcup_{i\in I}V_i$. So $\epsilon_i(v_i)$ is such a function, and can thus be applied to an element $j$ of $I$. When we do this, we write it as $\epsilon_i(v_i)(j)$. It can help avoid confusion to write this as $\left(\epsilon_i(v_i)\right)(j)$

 

[Math Amateur]

We are given that $spt(f)=\{i_1,...,i_t\}$. That means that $\exists v_{i_1}\in V_{i_1},...,v_{i_t}\in V_{i_t}$ such that
$$f=\epsilon_{i_1}(v_{i_1})+...+\epsilon_{i_t}(v_{i_t})$$

Then for $k\in \{1,...,t\}$ we have, by replacing $f$ by $\epsilon_{i}(v_{i})$ in the definition of $G(f)$:

$$G\circ\epsilon_{i}(v_{i})\equiv G(\epsilon_{i}(v_{i}))\equiv\sum_{j=1}^t g_{i_j}(\epsilon_{i}(v_{i})(i_j))$$

In your earlier thread $\epsilon_k$ was defined as the function from $V_k$ to the direct product $V$ that maps $v_k$ to
$(0,0,...0,v_k,0,...,0)$ where $v_k$ is in the $k$th position. That only covers finite direct sums. However it looks from the above as though Cooperstein was - between the two sections you quoted - moved on to defining and allowing infinite direct sums (because of his use of an index set $I$ of unspecified cardinality, rather than just labelling the component spaces as $V_1$ to $V_n$). That means that $\epsilon_k$ needs an appropriately modified definition. I'm guessing the definition he's using is something like that $\epsilon_i:V_i\to V$ such that $\epsilon_i(v_i)(j)$ is zero for all $j\in I$ except $i$, for which it gives $v_i$.

Applying that to the above equation, we have

$$G\circ\epsilon_{i}(v_{i})\equiv \sum_{j=1}^t g_{i_j}(\epsilon_{i}(v_{i})(i_j))
=g_{i}(v_i)$$
as required.

Hi Andrew ... thanks again for the help ...

But ... just a clarification ... ...

You write:

"... ...
We are given that $spt(f)=\{i_1,...,i_t\}$. That means that $\exists v_{i_1}\in V_{i_1},...,v_{i_t}\in V_{i_t}$ such that
$$f=\epsilon_{i_1}(v_{i_1})+...+\epsilon_{i_t}(v_{i_t})$$ ... ... "

and you follow this by writing ... :

" ... ... Then for $k\in \{1,...,t\}$ we have, by replacing $f$ by $\epsilon_{i}(v_{i})$ in the definition of $G(f)$:"


I do not follow this ... shouldn't you be replacing f by $$\epsilon_{i_1}(v_{i_1})+...+\epsilon_{i_t}(v_{i_t})$$ ... ... ... ?

Can you explain ...

Peter

 

[andrewkirk]

you follow this by writing ... :

" ... ... Then for $k\in \{1,...,t\}$ we have, by replacing $f$ by $\epsilon_{i}(v_{i})$ in the definition of $G(f)$:"


I do not follow this ... shouldn't you be replacing f by $$\epsilon_{i_1}(v_{i_1})+...+\epsilon_{i_t}(v_{i_t})$$ ... ... ... ?

In the definition of $G$, the symbol $f$ stands for any arbitrary element of $V$. Now $\epsilon_i(v_i)$ is such an element and thus can be slotted in as the argument to $G$ in that definition. Perhaps I should have added that $v_i$ is an arbitrary element of $V_i$. Note that the support of $\epsilon_i(v_i):I\to \bigcup_{i\in I}V_i$ is the singleton $\{i\}$, so the sum in the RHS of the definition of $G$ only has one element when the argument is $\epsilon_i(v_i)$, so we can discard the summation symbol.

I think Cooperstein has confused the issue by defining $f$ before he defines $G$, and thereby implying that $G$ somehow depends on $f$, which it doesn't! It would be better if he had written the following instead:

Define function $G:V\to W$ such that, $\forall f\in V$, $G(f)\equiv \sum_{j\in spt(f)} g_j(f(j))$.

Applying that to $\epsilon_i(v_i)$ then gives

$$G(\epsilon_i(v_i))\equiv \sum_{j\in spt(\epsilon_i(v_i))} g_j((\epsilon_i(v_i))(j))
=\sum_{j\in \{i\}} g_j((\epsilon_i(v_i))(j))
=g_i((\epsilon_i(v_i))(i))
=g_i(v_i)
$$

 

[Math Amateur]

In the definition of $G$, the symbol $f$ stands for any arbitrary element of $V$. Now $\epsilon_i(v_i)$ is such an element and thus can be slotted in as the argument to $G$ in that definition. Perhaps I should have added that $v_i$ is an arbitrary element of $V_i$. Note that the support of $\epsilon_i(v_i):I\to \bigcup_{i\in I}V_i$ is the singleton $\{i\}$, so the sum in the RHS of the definition of $G$ only has one element when the argument is $\epsilon_i(v_i)$, so we can discard the summation symbol.

I think Cooperstein has confused the issue by defining $f$ before he defines $G$, and thereby implying that $G$ somehow depends on $f$, which it doesn't! It would be better if he had written the following instead:

Define function $G:V\to W$ such that, $\forall f\in V$, $G(f)\equiv \sum_{j\in spt(f)} g_j(f(j))$.

Applying that to $\epsilon_i(v_i)$ then gives

$$G(\epsilon_i(v_i))\equiv \sum_{j\in spt(\epsilon_i(v_i))} g_j((\epsilon_i(v_i))(j))
=\sum_{j\in \{i\}} g_j((\epsilon_i(v_i))(j))
=g_i((\epsilon_i(v_i))(i))
=g_i(v_i)
$$

Hi Andrew,

Thanks to your posts I now have a much better understanding of what is going on ...

But just one further (minor) issue ...

You write:

" ... ... In the definition of $G$, the symbol $f$ stands for any arbitrary element of $V$. Now $\epsilon_i(v_i)$ is such an element and thus can be slotted in as the argument to $G$ in that definition. ... ... "

But ... is $\epsilon_i(v_i)$ really an arbitrary element? ... ... it has the special form of being an element with support equal to a one element set ... shouldn't we be taking a general element - that is an element with support equal to an n-element set where n is any integer ... ...

Can you help clarify this issue?

Peter

 

[andrewkirk]

You write:

" ... ... In the definition of $G$, the symbol $f$ stands for any arbitrary element of $V$. Now $\epsilon_i(v_i)$ is such an element and thus can be slotted in as the argument to $G$ in that definition. ... ... "

But ... is $\epsilon_i(v_i)$ really an arbitrary element? ... ... it has the special form of being an element with support equal to a one element set ... shouldn't we be taking a general element - that is an element with support equal to an n-element set where n is any integer ... ...

$f$ stands for the arbitrary element of $V$, not $\epsilon_i(v_i)$. The latter is a specific element.

What we are doing is substituting a specific element of $V$ for the arbitrary element $f$, in the first-order logical formula:

$$\forall f\in V:\ G(f)= \sum_{j\in spt(f)} g_j(f(j))$$

This is the type of operation enabled by the axiom schema of substitution (aka instantiation) which is Q5 in this axiomatisation of first order logic. The universal quantifier $\forall$ is what enables this substitution.

It's the same as if we take the formula $\forall x\in\mathbb{R}:\ x^2\geq 0$ and substitute the specific element -2 for the arbitrary element $x$. That gives us the valid formula $(-2)^2\geq 0$.