MHB Vector Spaces and Linear Transformations - Cooperstein Theorem 2.7

[Math Amateur]

I am reading Bruce Cooperstein's book: Advanced Linear Algebra ... ...

I am focused on Section 2.1 Introduction to Linear Transformations ... ...

I need help with understanding Theorem 2.7 ...

Theorem 2.7, its proof and some remarks read as follows:View attachment 5157I am having considerable trouble understanding what this Theorem is about ... can someone please explain the Theorem perhaps with a simple tangible example ...

My apologies for not asking a specific question ... but i am somewhat confused as to what this Theorem is saying ...

Hope someone can help ...

Peter*** EDIT ***

I also have no idea what Cooperstein is saying when, after presenting the Theorem and its proof he writes ...

" ... V is universal among all pairs (f, W) ... "

 

[Deveno]

Let's say that $W$ is a vector space (ANY vector space, over the same field as $V$). Furthermore, suppose that $f:\mathcal{B} \to W$ is *any* function.

To say that $V$ is universal among all pairs $(f,W)$ (and actually to be "precise" we mean the pair $(1_{\mathcal{B}},V)$ where $1_{\mathcal{B}}: \mathcal{B} \to V$ is the inclusion function:

$1_{\mathcal{B}}(v_j) = v_j$, for each $v_j \in \mathcal{B}$)

means that there exists a unique linear transformation: $T:V \to W$ such that:

$f = T \circ 1_{\mathcal{B}}$.

Explicitly:

$T\left(\sum\limits_j a_jv_j\right) = \sum\limits_j a_jf(v_j)$

This process is called "extending by linearity", or "defining a linear map by its action on a basis".

So, here is how it works in practice:

Let's use a simple example, we will take $F = \Bbb R$, and $V = \Bbb R^3$, and $\mathcal{B} = \{(1,0,0),(0,1,0),(0,0,1)\}$

For $W$, we will use $\Bbb R^2$ (we could use other examples, $\Bbb R^5$ would work, too, if we desired).

Now to determine $T$, we need a PAIR, $(f,W)$, so I need to say what $f$ is. Let's just pick $3$ elements of $W$ more or less at random, for the images of $(1,0,0),(0,1,0)$ and $(0,0,1)$:

$f((1,0,0)) = (2,4)$
$f((0,1,0)) = (-1,5)$
$f((0,0,1)) = (0,3)$.

My claim is now there is only ONE linear map $T:\Bbb R^3 \to \Bbb R^2$ that will satisfy:

$f = T \circ 1_{\mathcal{B}}$.

To know what $T$ might be, it suffices to define it for an arbitrary $(x,y,z) \in \Bbb R^3$, that is we must specify what $T(x,y,z))$ is equal to.

Now $T((x,y,z)) = T((x,0,0) + (0,y,0) + (0,0,z)) = T(x(1,0,0) + y(0,1,0) + z(0,0,1))$.

If $T$ is to be linear, we must have:

$T(x(1,0,0) + y(0,1,0) + z(0,0,1)) = xT((1,0,0)) + yT((0,1,0)) + zT((0,0,1))$

$= x(2,4) + y(-1,5) + z(0,3) = (2x,4x) + (-y,5y) + (0,3z) = (2x-y,4x+5y+3z)$

So if such a linear $T$ exists, we have to have:

$T((x,y,z)) = (2x-y,4x+5y+3z)$

This shows that if $T$ exists, it must be unique. So all that remains (which I leave to you) is to show that the $T$ defined above, *is indeed linear*, which shows existence.

There is a similarity here with the universal mapping property of a free group, the mapping $1_{\mathcal{B}}$ here corresponds to *inclusion of generators* of a set $X$ into the free group generated by $X$-this is no accident:

Vector spaces are free $F$-modules, and any two bases are set-isomorphic (a concept enshrined as the definition of "dimension").

Now this is a rather abstract approach-the usual way vector spaces are introduced is as LINEAR COMBINATIONS of basis elements. "Formal" linear combinations are what you have to have, in order to have CLOSURE of a set under addition and scalar multiplication.

There is one thing to be careful of, here: "formal linear combinations" assume beforehand, that the set-elements are linearly independent (often just by declaring them so), that is, no actual algebraic relations are held to hold between the set-elements we take as basis elements. But if we are dealing with set-elements that already have some algebraic relationship (such as points in the plane, or $\Bbb R \oplus \Bbb R$) we cannot assume this. That is why it is *key* in this theorem of Cooperstein's, that $\mathcal{B}$ be a *basis*, and not just some set of vectors.

The long and short of this is, if you have a basis, like say, $\{1,x,x^2,x^3,\dots\}$ for $\Bbb R[x]$, then any linear map $T: \Bbb R[x] \to W$ can be defined *just by looking at $T(1),T(x),T(x^2),T(x^3),\dots$ etc.*

The usual way this is presented, is by considering *matrix representations*. If the basis $\mathcal{B}$ is given, we can represent the linear combination:

$v = a_1v_1 +\cdots + a_nv_n$ by the $n \times 1$ array (column vector):

$\begin{bmatrix}a_1\\ \vdots\\a_n\end{bmatrix}$ (note how this suppresses the basis as "understood"). Sometimes this is written $[v]_{\mathcal{B}}$.

In this basis, if we know $T(v_j)$, then that is the same as knowing:

$[T]_{\mathcal{B}'}^{\mathcal{B}}\begin{bmatrix}0\\ \vdots\\1\\ \vdots\\0\end{bmatrix}$ for some representation of the $T(v_j)$ in a basis $\mathcal{B}'$ for $W$.

Now what the matrix on the right "does" to the matrix form of $T$, is pick out the $j$-th column, that is:

$[T]_{\mathcal{B}'}^{\mathcal{B}} = \begin{bmatrix}|&|&\cdots&|\\ [T(v_1)]_{\mathcal{B}'}&T(v_2)]_{\mathcal{B}'}&
\cdots&[T(v_n)]_{\mathcal{B}'} \\|&|&\cdots&|\end{bmatrix}$

For example, in my illustration above, the matrix for $T$ is seen to be:

$[T]_{\mathcal{B}'}^{\mathcal{B}} = \begin{bmatrix}2&-1&0\\4&5&3\end{bmatrix}$

if we use the basis $\mathcal{B}' = \{(1,0),(0,1)\}$ for $\Bbb R^2$.

Hopefully, it should be clear this is the ONLY 3x2 matrix that maps:

$\begin{bmatrix}1\\0\\0\end{bmatrix} \mapsto \begin{bmatrix}2\\4\end{bmatrix}$

$\begin{bmatrix}0\\1\\0\end{bmatrix} \mapsto \begin{bmatrix}-1\\5\end{bmatrix}$

$\begin{bmatrix}0\\0\\1\end{bmatrix} \mapsto \begin{bmatrix}0\\3\end{bmatrix}$

via matrix multiplication from the left.

 

[Math Amateur]

Let's say that $W$ is a vector space (ANY vector space, over the same field as $V$). Furthermore, suppose that $f:\mathcal{B} \to W$ is *any* function.

To say that $V$ is universal among all pairs $(f,W)$ (and actually to be "precise" we mean the pair $(1_{\mathcal{B}},V)$ where $1_{\mathcal{B}}: \mathcal{B} \to V$ is the inclusion function:

$1_{\mathcal{B}}(v_j) = v_j$, for each $v_j \in \mathcal{B}$)

means that there exists a unique linear transformation: $T:V \to W$ such that:

$f = T \circ 1_{\mathcal{B}}$.

Explicitly:

$T\left(\sum\limits_j a_jv_j\right) = \sum\limits_j a_jf(v_j)$

This process is called "extending by linearity", or "defining a linear map by its action on a basis".

So, here is how it works in practice:

Let's use a simple example, we will take $F = \Bbb R$, and $V = \Bbb R^3$, and $\mathcal{B} = \{(1,0,0),(0,1,0),(0,0,1)\}$

For $W$, we will use $\Bbb R^2$ (we could use other examples, $\Bbb R^5$ would work, too, if we desired).

Now to determine $T$, we need a PAIR, $(f,W)$, so I need to say what $f$ is. Let's just pick $3$ elements of $W$ more or less at random, for the images of $(1,0,0),(0,1,0)$ and $(0,0,1)$:

$f((1,0,0)) = (2,4)$
$f((0,1,0)) = (-1,5)$
$f((0,0,1)) = (0,3)$.

My claim is now there is only ONE linear map $T:\Bbb R^3 \to \Bbb R^2$ that will satisfy:

$f = T \circ 1_{\mathcal{B}}$.

To know what $T$ might be, it suffices to define it for an arbitrary $(x,y,z) \in \Bbb R^3$, that is we must specify what $T(x,y,z))$ is equal to.

Now $T((x,y,z)) = T((x,0,0) + (0,y,0) + (0,0,z)) = T(x(1,0,0) + y(0,1,0) + z(0,0,1))$.

If $T$ is to be linear, we must have:

$T(x(1,0,0) + y(0,1,0) + z(0,0,1)) = xT((1,0,0)) + yT((0,1,0)) + zT((0,0,1))$

$= x(2,4) + y(-1,5) + z(0,3) = (2x,4x) + (-y,5y) + (0,3z) = (2x-y,4x+5y+3z)$

So if such a linear $T$ exists, we have to have:

$T((x,y,z)) = (2x-y,4x+5y+3z)$

This shows that if $T$ exists, it must be unique. So all that remains (which I leave to you) is to show that the $T$ defined above, *is indeed linear*, which shows existence.

There is a similarity here with the universal mapping property of a free group, the mapping $1_{\mathcal{B}}$ here corresponds to *inclusion of generators* of a set $X$ into the free group generated by $X$-this is no accident:

Vector spaces are free $F$-modules, and any two bases are set-isomorphic (a concept enshrined as the definition of "dimension").

Now this is a rather abstract approach-the usual way vector spaces are introduced is as LINEAR COMBINATIONS of basis elements. "Formal" linear combinations are what you have to have, in order to have CLOSURE of a set under addition and scalar multiplication.

There is one thing to be careful of, here: "formal linear combinations" assume beforehand, that the set-elements are linearly independent (often just by declaring them so), that is, no actual algebraic relations are held to hold between the set-elements we take as basis elements. But if we are dealing with set-elements that already have some algebraic relationship (such as points in the plane, or $\Bbb R \oplus \Bbb R$) we cannot assume this. That is why it is *key* in this theorem of Cooperstein's, that $\mathcal{B}$ be a *basis*, and not just some set of vectors.

The long and short of this is, if you have a basis, like say, $\{1,x,x^2,x^3,\dots\}$ for $\Bbb R[x]$, then any linear map $T: \Bbb R[x] \to W$ can be defined *just by looking at $T(1),T(x),T(x^2),T(x^3),\dots$ etc.*

The usual way this is presented, is by considering *matrix representations*. If the basis $\mathcal{B}$ is given, we can represent the linear combination:

$v = a_1v_1 +\cdots + a_nv_n$ by the $n \times 1$ array (column vector):

$\begin{bmatrix}a_1\\ \vdots\\a_n\end{bmatrix}$ (note how this suppresses the basis as "understood"). Sometimes this is written $[v]_{\mathcal{B}}$.

In this basis, if we know $T(v_j)$, then that is the same as knowing:

$[T]_{\mathcal{B}'}^{\mathcal{B}}\begin{bmatrix}0\\ \vdots\\1\\ \vdots\\0\end{bmatrix}$ for some representation of the $T(v_j)$ in a basis $\mathcal{B}'$ for $W$.

Now what the matrix on the right "does" to the matrix form of $T$, is pick out the $j$-th column, that is:

$[T]_{\mathcal{B}'}^{\mathcal{B}} = \begin{bmatrix}|&|&\cdots&|\\ [T(v_1)]_{\mathcal{B}'}&T(v_2)]_{\mathcal{B}'}&
\cdots&[T(v_n)]_{\mathcal{B}'} \\|&|&\cdots&|\end{bmatrix}$

For example, in my illustration above, the matrix for $T$ is seen to be:

$[T]_{\mathcal{B}'}^{\mathcal{B}} = \begin{bmatrix}2&-1&0\\4&5&3\end{bmatrix}$

if we use the basis $\mathcal{B}' = \{(1,0),(0,1)\}$ for $\Bbb R^2$.

Hopefully, it should be clear this is the ONLY 3x2 matrix that maps:

$\begin{bmatrix}1\\0\\0\end{bmatrix} \mapsto \begin{bmatrix}2\\4\end{bmatrix}$

$\begin{bmatrix}0\\1\\0\end{bmatrix} \mapsto \begin{bmatrix}-1\\5\end{bmatrix}$

$\begin{bmatrix}0\\0\\1\end{bmatrix} \mapsto \begin{bmatrix}0\\3\end{bmatrix}$

via matrix multiplication from the left.

Thanks so much for the support in helping me to understand this Theorem and its implications ...

I will be working carefully through the detail of your post shortly ...

Thanks again for the extensive help ... it is much appreciated ...

Peter

 

[Math Amateur]

Let's say that $W$ is a vector space (ANY vector space, over the same field as $V$). Furthermore, suppose that $f:\mathcal{B} \to W$ is *any* function.

To say that $V$ is universal among all pairs $(f,W)$ (and actually to be "precise" we mean the pair $(1_{\mathcal{B}},V)$ where $1_{\mathcal{B}}: \mathcal{B} \to V$ is the inclusion function:

$1_{\mathcal{B}}(v_j) = v_j$, for each $v_j \in \mathcal{B}$)

means that there exists a unique linear transformation: $T:V \to W$ such that:

$f = T \circ 1_{\mathcal{B}}$.

Explicitly:

$T\left(\sum\limits_j a_jv_j\right) = \sum\limits_j a_jf(v_j)$

This process is called "extending by linearity", or "defining a linear map by its action on a basis".

So, here is how it works in practice:

Let's use a simple example, we will take $F = \Bbb R$, and $V = \Bbb R^3$, and $\mathcal{B} = \{(1,0,0),(0,1,0),(0,0,1)\}$

For $W$, we will use $\Bbb R^2$ (we could use other examples, $\Bbb R^5$ would work, too, if we desired).

Now to determine $T$, we need a PAIR, $(f,W)$, so I need to say what $f$ is. Let's just pick $3$ elements of $W$ more or less at random, for the images of $(1,0,0),(0,1,0)$ and $(0,0,1)$:

$f((1,0,0)) = (2,4)$
$f((0,1,0)) = (-1,5)$
$f((0,0,1)) = (0,3)$.

My claim is now there is only ONE linear map $T:\Bbb R^3 \to \Bbb R^2$ that will satisfy:

$f = T \circ 1_{\mathcal{B}}$.

To know what $T$ might be, it suffices to define it for an arbitrary $(x,y,z) \in \Bbb R^3$, that is we must specify what $T(x,y,z))$ is equal to.

Now $T((x,y,z)) = T((x,0,0) + (0,y,0) + (0,0,z)) = T(x(1,0,0) + y(0,1,0) + z(0,0,1))$.

If $T$ is to be linear, we must have:

$T(x(1,0,0) + y(0,1,0) + z(0,0,1)) = xT((1,0,0)) + yT((0,1,0)) + zT((0,0,1))$

$= x(2,4) + y(-1,5) + z(0,3) = (2x,4x) + (-y,5y) + (0,3z) = (2x-y,4x+5y+3z)$

So if such a linear $T$ exists, we have to have:

$T((x,y,z)) = (2x-y,4x+5y+3z)$

This shows that if $T$ exists, it must be unique. So all that remains (which I leave to you) is to show that the $T$ defined above, *is indeed linear*, which shows existence.

There is a similarity here with the universal mapping property of a free group, the mapping $1_{\mathcal{B}}$ here corresponds to *inclusion of generators* of a set $X$ into the free group generated by $X$-this is no accident:

Vector spaces are free $F$-modules, and any two bases are set-isomorphic (a concept enshrined as the definition of "dimension").

Now this is a rather abstract approach-the usual way vector spaces are introduced is as LINEAR COMBINATIONS of basis elements. "Formal" linear combinations are what you have to have, in order to have CLOSURE of a set under addition and scalar multiplication.

There is one thing to be careful of, here: "formal linear combinations" assume beforehand, that the set-elements are linearly independent (often just by declaring them so), that is, no actual algebraic relations are held to hold between the set-elements we take as basis elements. But if we are dealing with set-elements that already have some algebraic relationship (such as points in the plane, or $\Bbb R \oplus \Bbb R$) we cannot assume this. That is why it is *key* in this theorem of Cooperstein's, that $\mathcal{B}$ be a *basis*, and not just some set of vectors.

The long and short of this is, if you have a basis, like say, $\{1,x,x^2,x^3,\dots\}$ for $\Bbb R[x]$, then any linear map $T: \Bbb R[x] \to W$ can be defined *just by looking at $T(1),T(x),T(x^2),T(x^3),\dots$ etc.*

The usual way this is presented, is by considering *matrix representations*. If the basis $\mathcal{B}$ is given, we can represent the linear combination:

$v = a_1v_1 +\cdots + a_nv_n$ by the $n \times 1$ array (column vector):

$\begin{bmatrix}a_1\\ \vdots\\a_n\end{bmatrix}$ (note how this suppresses the basis as "understood"). Sometimes this is written $[v]_{\mathcal{B}}$.

In this basis, if we know $T(v_j)$, then that is the same as knowing:

$[T]_{\mathcal{B}'}^{\mathcal{B}}\begin{bmatrix}0\\ \vdots\\1\\ \vdots\\0\end{bmatrix}$ for some representation of the $T(v_j)$ in a basis $\mathcal{B}'$ for $W$.

Now what the matrix on the right "does" to the matrix form of $T$, is pick out the $j$-th column, that is:

$[T]_{\mathcal{B}'}^{\mathcal{B}} = \begin{bmatrix}|&|&\cdots&|\\ [T(v_1)]_{\mathcal{B}'}&T(v_2)]_{\mathcal{B}'}&
\cdots&[T(v_n)]_{\mathcal{B}'} \\|&|&\cdots&|\end{bmatrix}$

For example, in my illustration above, the matrix for $T$ is seen to be:

$[T]_{\mathcal{B}'}^{\mathcal{B}} = \begin{bmatrix}2&-1&0\\4&5&3\end{bmatrix}$

if we use the basis $\mathcal{B}' = \{(1,0),(0,1)\}$ for $\Bbb R^2$.

Hopefully, it should be clear this is the ONLY 3x2 matrix that maps:

$\begin{bmatrix}1\\0\\0\end{bmatrix} \mapsto \begin{bmatrix}2\\4\end{bmatrix}$

$\begin{bmatrix}0\\1\\0\end{bmatrix} \mapsto \begin{bmatrix}-1\\5\end{bmatrix}$

$\begin{bmatrix}0\\0\\1\end{bmatrix} \mapsto \begin{bmatrix}0\\3\end{bmatrix}$

via matrix multiplication from the left.

Thanks again for your help and support, Deveno ... but i need some further clarification and help ...You write:

"... ... To say that $V$ is universal among all pairs $(f,W)$ (and actually to be "precise" we mean the pair $(1_{\mathcal{B}},V)$ where $1_{\mathcal{B}}: \mathcal{B} \to V$ is the inclusion function:

$1_{\mathcal{B}}(v_j) = v_j$, for each $v_j \in \mathcal{B}$)

means that there exists a unique linear transformation: $T:V \to W$ such that:

$f = T \circ 1_{\mathcal{B}}$.

Explicitly:

$T\left(\sum\limits_j a_jv_j\right) = \sum\limits_j a_jf(v_j)$

This process is called "extending by linearity", or "defining a linear map by its action on a basis".
I need some help in order to fully understand this ... in particular I need clarification of the role of $$1_\mathcal{B}$$ ...My problem is that it appears that $$1_\mathcal{B}$$ 'does nothing' when we are mapping an element $$v \in V$$, say $$v = \sum a_j v_j$$ where the $$v_j$$ are the basis vectors ... ...
To demonstrate ... as I see it ...If we consider $$v = \sum a_j f(v_j)$$ we have ...$$\sum a_j f(v_j)

= \sum a_j [ T \circ 1_{\mathcal{B}} (v_j) ]

= \sum a_j [ T ( 1_{\mathcal{B}} (v_j) ) ]

= \sum a_j [ T (v_j) ]$$ ... ... since $$1_{\mathcal{B}} (v_j) = v_j
$$... so ... ? ... the mapping $$1_{\mathcal{B}}$$ seems redundant ... that is it does not, as far as I can see, 'do anything' ... ?

... that is when we write $$\sum a_j f(v_j)$$ we might just as well write $$\sum a_j [ T (v_j) ]$$ ... ... ?
To try to make my problem/confusion clearer ... consider $$T ( \sum a_j v_j )$$ where the $$v_j$$ are the basis vectors of V ... ...

... we have ...

$$T ( \sum a_j v_j ) = \sum a_j [ T ( v_j ) ] = \sum a_j f (v_j)$$ ... ... So, again the $$1_{\mathcal{B}}$$ seems redundant ... ... BUT ... ... I suspect that I am not interpreting things correctly ...
Can you help to clarify my issue ...

Peter*** EDIT ***

I have the vague idea that the central thought in this matter is that every vector of V can be expressed in terms of the basis $$\mathcal{B}$$ ... and so if we know how $$\mathcal{B}$$ is mapped under T then we know how any vector in $$V$$ is mapped under $$T$$ ... so, somehow we want to work from the mapping of $$\mathcal{B}$$ under $$T$$ ... and so we use the inclusion map to express the fact that we are only concerned with how the basis is mapped under $$T$$ ...

... something like that anyway ... must go back and check your example ... hmmm... ? ... still reflecting ...

Peter

 

[Deveno]

Inclusions are spectacularly dull mappings. I'll not argue that.

And yes, you are correct, the gist of what is happening is that we can extend a basis uniquely to its span, by considering for any element in the span the unique linear combination of basis elements that the span element is.

But you are missing something kind of subtle, so subtle it's hard to notice.

When we talk about basis elements, and linear combinations, our focus is on individual vectors, such as:

$\sum\limits_j a_jv_j$.

When we write:

$\exists ! T \in \text{Hom}_F(V,W): f = T \circ 1_{\mathcal{B}}$

we are no longer concerned with individual elements, our focus is entirely on MAPPINGS.

Consider the analogy with free groups:

given a set $X$, the free group generated by $X$ is universal among all pairs $(f,G)$ where $f:X \to G$ is any function, and $G$ is any group.

What this means is that given the pair $(f,G)$, there exists a unique group homomorphism $F(X) \to G$ that maps generators (of $F(X)$) to the images of $f$. If $f(X)$ is a generating set for $G$, we call this "mapping generators to generators":

$\phi: F(X) \to G$ is defined by: $\phi(x_i) = f(x_i) = f \circ \iota (x_i)$, where $\iota: X \to F[X]$ is the canonical inclusion: "inclusion of generators" (it sends $x_i$ to the "word" $x_i$).

As an example of how this works, suppose $X = \{x,y\}$ so that $F(X) = F_2$, the free group on two generators.

Consider $D_4 = \langle r,s \rangle$, the dihedral group of order 8.

We have the obvious set-mapping:

$x \mapsto r$
$y \mapsto s$, and so we can define:

$\phi: F_2 \to D_4$ by $\phi(x) = r$, and $\phi(y) = s$.

Since $\{r,s\}$ is a generating set for $D_4$, this homomorphism is clearly onto. Thus $D_4$ is a *homomorphic image* (quotient group) of $F_2$.

It is instructive to consider $\text{ker }\phi$. Note that $r^4 = 1$, so the kernel must contain $\langle x^4\rangle = \langle xxxx \rangle$ (so any word with 4 $x$'s in a row maps to a similar word (in $D_4$) with those 4 $x$'s replaced by the image of the empty word, $1$). Similarly, the kernel also contains $\langle y^2\rangle$.

Finally, we have: $sr = r^{-1}s$, that is $(sr)^2 = 1$, so the kernel also contains $\langle yxyx\rangle$.

This is enough to *completely* determine $D_4$, so the kernel is the smallest normal subgroup of $F_2$ containing:

$\{x^4,y^2,yxyx\}$, which turns out to be all conjugates of any element of $\langle x^4,y^2,yxyx\rangle$. For example:

$xyxyyxxxxxyxyx^{-1}y^{-1}x^{-1} \mapsto 1$.

This allows us to DEFINE:

$D_4 = \langle r,s: r^4 = s^2 = 1, sr = r^{-1}s\rangle$

rather than explicitly listing a large Cayley table.

In vector spaces, the generating set is replaced by a basis (a spanning set, while also a generating set, does not "freely" generate a vector space $V$ if it is linearly dependent, because linear dependence IS a non-trivial relation between the generators).

Cooperstein, in his preface to this theorem, alludes to the usefulness of considering UMP's instead of "explicit" constructions when he talks about tensor products. In point of fact, tensor products are very difficult to construct "explicitly", to construct $V \otimes W$,for two vector spaces for example, one has to consider a very unwieldy quotient space of $V \times W$, whereas using the UMP one can consider it as a construction that turns *bilinear maps* from $V \times W$ into linear maps from $V \otimes W$, which let's one get results more easily than digging through the often complicated cosets.

In fact, UMP's are the "preferred tools" for proving a lot of algebraic theorems. Here is another example: one makes quotient objects (quotient groups, rings, algebras, vector spaces, modules) in order to annihilate (send to the zero-object) some subobject. The quotient object $A/B$ is thus seen to satisfy a universal mapping property: it is universal among all pairs $(p,C)$ of homomorphisms $p: A \to C$ that annihilate $B$.

This property alone allows us to recapture the fundamental homomorphism theorem (although I will not prove that here).

The benefits of such an approach (in vector spaces, as you are studying now) is that you can prove a lot of things "just from linearity". In other words, *portable* proofs, which do not depend on the specific ways you defined a particular vector space, or the specific elements you are calculating with, or the particular basis you happened to choose. It's good to know these things, because often we wish to choose a basis which makes our calculations EASY (for example, using the basis: $\{x^n: n \in \Bbb N\}$ for polynomials which allows us to use coefficients as coordinates).

 

[Math Amateur]

Inclusions are spectacularly dull mappings. I'll not argue that.

And yes, you are correct, the gist of what is happening is that we can extend a basis uniquely to its span, by considering for any element in the span the unique linear combination of basis elements that the span element is.

But you are missing something kind of subtle, so subtle it's hard to notice.

When we talk about basis elements, and linear combinations, our focus is on individual vectors, such as:

$\sum\limits_j a_jv_j$.

When we write:

$\exists ! T \in \text{Hom}_F(V,W): f = T \circ 1_{\mathcal{B}}$

we are no longer concerned with individual elements, our focus is entirely on MAPPINGS.

Consider the analogy with free groups:

given a set $X$, the free group generated by $X$ is universal among all pairs $(f,G)$ where $f:X \to G$ is any function, and $G$ is any group.

What this means is that given the pair $(f,G)$, there exists a unique group homomorphism $F(X) \to G$ that maps generators (of $F(X)$) to the images of $f$. If $f(X)$ is a generating set for $G$, we call this "mapping generators to generators":

$\phi: F(X) \to G$ is defined by: $\phi(x_i) = f(x_i) = f \circ \iota (x_i)$, where $\iota: X \to F[X]$ is the canonical inclusion: "inclusion of generators" (it sends $x_i$ to the "word" $x_i$).

As an example of how this works, suppose $X = \{x,y\}$ so that $F(X) = F_2$, the free group on two generators.

Consider $D_4 = \langle r,s \rangle$, the dihedral group of order 8.

We have the obvious set-mapping:

$x \mapsto r$
$y \mapsto s$, and so we can define:

$\phi: F_2 \to D_4$ by $\phi(x) = r$, and $\phi(y) = s$.

Since $\{r,s\}$ is a generating set for $D_4$, this homomorphism is clearly onto. Thus $D_4$ is a *homomorphic image* (quotient group) of $F_2$.

It is instructive to consider $\text{ker }\phi$. Note that $r^4 = 1$, so the kernel must contain $\langle x^4\rangle = \langle xxxx \rangle$ (so any word with 4 $x$'s in a row maps to a similar word (in $D_4$) with those 4 $x$'s replaced by the image of the empty word, $1$). Similarly, the kernel also contains $\langle y^2\rangle$.

Finally, we have: $sr = r^{-1}s$, that is $(sr)^2 = 1$, so the kernel also contains $\langle yxyx\rangle$.

This is enough to *completely* determine $D_4$, so the kernel is the smallest normal subgroup of $F_2$ containing:

$\{x^4,y^2,yxyx\}$, which turns out to be all conjugates of any element of $\langle x^4,y^2,yxyx\rangle$. For example:

$xyxyyxxxxxyxyx^{-1}y^{-1}x^{-1} \mapsto 1$.

This allows us to DEFINE:

$D_4 = \langle r,s: r^4 = s^2 = 1, sr = r^{-1}s\rangle$

rather than explicitly listing a large Cayley table.

In vector spaces, the generating set is replaced by a basis (a spanning set, while also a generating set, does not "freely" generate a vector space $V$ if it is linearly dependent, because linear dependence IS a non-trivial relation between the generators).

Cooperstein, in his preface to this theorem, alludes to the usefulness of considering UMP's instead of "explicit" constructions when he talks about tensor products. In point of fact, tensor products are very difficult to construct "explicitly", to construct $V \otimes W$,for two vector spaces for example, one has to consider a very unwieldy quotient space of $V \times W$, whereas using the UMP one can consider it as a construction that turns *bilinear maps* from $V \times W$ into linear maps from $V \otimes W$, which let's one get results more easily than digging through the often complicated cosets.

In fact, UMP's are the "preferred tools" for proving a lot of algebraic theorems. Here is another example: one makes quotient objects (quotient groups, rings, algebras, vector spaces, modules) in order to annihilate (send to the zero-object) some subobject. The quotient object $A/B$ is thus seen to satisfy a universal mapping property: it is universal among all pairs $(p,C)$ of homomorphisms $p: A \to C$ that annihilate $B$.

This property alone allows us to recapture the fundamental homomorphism theorem (although I will not prove that here).

The benefits of such an approach (in vector spaces, as you are studying now) is that you can prove a lot of things "just from linearity". In other words, *portable* proofs, which do not depend on the specific ways you defined a particular vector space, or the specific elements you are calculating with, or the particular basis you happened to choose. It's good to know these things, because often we wish to choose a basis which makes our calculations EASY (for example, using the basis: $\{x^n: n \in \Bbb N\}$ for polynomials which allows us to use coefficients as coordinates).

Hi Deveno,

Thanks for the support and help ... your significant help is much appreciated ...

Working through the detail of your post now ...

Peter