Xjyuk

The example is given below:

But I do not understand the details of calculating $phi_BB(alpha_v)$, could anyone explain this for me please?

The definition of $phi_BB(alpha_v)$ is given below:

EDIT:
I mean how the definition of the linear transformation given affect the matrix?

Part of the problem is that Proposition 8.1 is not a definition. It doesn't tell you what $Phi_BD$ is, or how to compute it. It simply asserts existence.

It's also not particularly well-stated as a proposition, since it asserts the existence of a family of isomorphisms based on pairs of bases $(B, D)$ on $V$ and $W$ respectively, but doesn't specify any way in which said isomorphisms differ. If you could find just one (out of the infinitely many) isomorphisms between $operatornameHom(V, W)$ and $M_k times n(F)$ (call it $phi$), then letting $Phi_BD = phi$ would technically satisfy the proposition, and constitute a proof!

Fortunately, I do know what the proposition is getting at. There is a very natural map $Phi_BD$, taking a linear map $alpha : V to W$, to a $k times n$ matrix.

The fundamental, intuitive idea behind this map is the idea that linear maps are entirely determined by their action on a basis. Let's say you have a linear map $alpha : V to W$, and a basis $B = (v_1, ldots, v_n)$ of $V$. That is, every vector $v in V$ can be expressed uniquely as a linear combination of the vectors $v_1, ldots, v_n$. If we know the values of $alpha(v_1), ldots, alpha(v_n)$, then we essentially know the value of $alpha(v)$ for any $v$, through linearity. The process involves first finding the unique $a_1, ldots, a_n in F$ such that
$$v = a_1 v_1 + ldots + a_n v_n.$$
Then, using linearity,
$$alpha(v) = alpha(a_1 v_1 + ldots + a_n v_n) = a_1 alpha(v_1) + ldots + a_n alpha(v_n).$$

As an example of this principle in action, let's say that you had a linear map $alpha : BbbR^2 to BbbR^3$, and all you knew about $alpha$ was that $alpha(1, 1) = (2, -1, 1)$ and $alpha(1, -1) = (0, 0, 4)$. What would be the value of $alpha(2, 4)$?

To solve this, first express
$$(2, 4) = 3(1, 1) + 1(1, -1)$$
(note that this linear combination is unique, since $((1, 1), (1, -1))$ is a basis for $BbbR^2$, and we could have done something similar for any vector, not just $(2, 4)$). Then,
$$alpha(2, 4) = 3alpha(1, 1) + 1 alpha(1, -1) = 3(2, -1, 1) + 1(0, 0, 4) = (6, -3, 7).$$
There is a converse to this principle too: if you start with a basis $(v_1, ldots, v_n)$ for $V$, and pick an arbitrary list of vectors $(w_1, ldots, w_n)$ from $W$ (not necessarily a basis), then there exists a unique linear transformation $alpha : V to W$ such that $alpha(v_i) = w_i$. So, you don't even need to assume an underlying linear transformation exists! Just map the basis vectors wherever you want in $W$, without restriction, and there will be a (unique) linear map that maps the basis in this way.

That is, if we fix a basis $B = (v_1, ldots, v_n)$ of $V$, then we can make a bijective correspondence between the linear maps from $V$ to $W$, and lists of $n$ vectors in $W$. The map
$$operatornameHom(V, W) to W^n : alpha mapsto (alpha(v_1), ldots, alpha(v_n))$$
is bijective. This is related to the $Phi$ maps, but we still need to go one step further.

Now, let's take a basis $D = (w_1, ldots, w_m)$ of $W$. That is, each vector in $W$ can be uniquely written as a linear combination of $w_1, ldots, w_m$. So, we have a natural map taking a vector
$$w = b_1 w_1 + ldots + b_n w_n$$
to its coordinate column vector
$$[w]_D = beginbmatrix b_1 \ vdots \ b_n endbmatrix.$$
This map is an isomorphism between $W$ and $F^m$; we lose no information if we choose to express vectors in $W$ this way.

So, if we can express linear maps $alpha : V to W$ as a list of vectors in $W$, we could just as easily write this list of vectors in $W$ as a list of coordinate column vectors in $F^m$. Instead of thinking about $(alpha(v_1), ldots, alpha(v_n))$, think about
$$([alpha(v_1)]_D, ldots, [alpha(v_n)]_D).$$
Equivalently, this list of $n$ column vectors could be thought of as a matrix:
$$left[beginarrayc & & \ [alpha(v_1)]_D & cdots & [alpha(v_n)]_D \ & & endarrayright].$$
This matrix is $Phi_BD$! The procedure can be summed up as follows:

1. Compute $alpha$ applied to each basis vector in $B$ (i.e. compute $alpha(v_1), ldots, alpha(v_n)$), then


2. Compute the coordinate column vector of each of these transformed vectors with respect to the basis $D$ (i.e. $[alpha(v_1)]_D, ldots, [alpha(v_n)]_D$), and finally,


3. Put these column vectors into a single matrix.

Note that step 2 typically takes the longest. For each $alpha(v_i)$, you need to find (somehow) the scalars $b_i1, ldots, b_im$ such that
$$alpha(v_i) = b_i1 w_1 + ldots + b_im w_m$$
where $D = (w_1, ldots, w_m)$ is the basis for $W$. How to solve this will depend on what $W$ consists of (e.g. $k$-tuples of real numbers, polynomials, matrices, functions, etc), but it will almost always reduce to solving a system of linear equations in the field $F$.

As for why we represent linear maps this way, I think you'd better read further in your textbook. It essentially comes down to the fact that, given any $v in V$,
$$[alpha(v)]_D = Phi_BD(alpha) cdot [v]_B,$$
which reduces the (potentially complex) process of applying an abstract linear transformation on an abstract vector $v in V$ down to simple matrix multiplication in $F$. I discuss this (with different notation) in this answer, but I suggest looking through your book first. Also, this answer has a nice diagram, but different notation again.

So, let's get into your example. In this case, $B = D = ((1, 0, 0), (0, 1, 0), (0, 0, 1))$, a basis for $V = W = BbbR^3$. We have a fixed vector $w = (w_1, w_2, w_3)$ (which is $v$ in the question, but I've chosen to change it to $w$ and keep $v$ as our dummy variable). Our linear map is $alpha_w : BbbR^3 to BbbR^3$ such that $alpha_w(v) = w times v$. Let's follow the steps.

First, we compute $alpha_w(1, 0, 0), alpha_w(0, 1, 0), alpha_w(0, 0, 1)$:
beginalign*
alpha_w(1, 0, 0) &= (w_1, w_2, w_3) times (1, 0, 0) = (0, w_3, -w_2) \
alpha_w(0, 1, 0) &= (w_1, w_2, w_3) times (0, 1, 0) = (-w_3, 0, w_1) \
alpha_w(0, 0, 1) &= (w_1, w_2, w_3) times (0, 0, 1) = (w_2, -w_1, 0).
endalign*

Second, we need to write these vectors as coordinate column vectors with respect to $B$. Fortunately, $B$ is the standard basis; we always have, for any $v = (a, b, c) in BbbR^3$,
$$(a, b, c) = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) implies [(a, b, c)]_B = beginbmatrix a \ b \ cendbmatrix.$$
In other words, we essentially just transpose these vectors to columns, giving us,
$$beginbmatrix 0 \ w_3 \ -w_2endbmatrix, beginbmatrix -w_3 \ 0 \ w_1endbmatrix, beginbmatrix w_2 \ -w_1 \ 0endbmatrix.$$

Last step: put these in a matrix:

$$Phi_BB(alpha_w) = beginbmatrix 0 & -w_3 & w_2 \ w_3 & 0 & -w_1 \ -w_2 & w_1 & 0 endbmatrix.$$

With the equations of $alpha_v$:

Let $:w=^mathrm tmkern-1.5mu(x, y,z)$. The coordinates of $vtimes w$ are obtained as the cofactors of the determinant (along the first row):

$$beginvmatrix
vec i&vec j&vec k \ a_1&a_2 & a_3 \ x&y&z
endvmatrix rightsquigarrow beginpmatrix
a_2z-a_3y\a_3x-a_1z \a_1y-a_2x
endpmatrix=beginpmatrix
0&-a_3&a_2\a_3& 0 &-a_1 \ -a_2 &a_1&0
endpmatrixbeginpmatrix
x \y\z
endpmatrix$$

The details probably come in the proof of Theorem 8.1 (which you should read).

Let $B = (v_1,dots,v_n)$ and $D = (w_1,dots,w_k)$ be the given bases. Suppose that $alphainoperatornameHom(V,W)$. For each $i$ in $1,dots,n$ there exist scalars $phi_ij in F$ such that
$$
alpha(v_i) = phi_1iw_1 + phi_2iw_2 + dots + phi_ki w_k
$$
Set $Phi_BD(alpha)$ to be the $ktimes n$ matrix whose $(i,j)$-th entry is $phi_ij$.

Now we come to angryavian's suggestion. Here $V = W = mathbbR^3$, and $B = D = (e_1,e_2,e_3)$. Moreover, $alpha(w) = v times w$ for a fixed $v = beginbmatrix a_1 \ a_2 \ a_3 endbmatrix$. So you need to find the coefficients of $alpha(e_1)$, $alpha(e_2)$ and $alpha(e_3)$ in the basis $(e_1,e_2,e_3)$.

The first column of the matrix is $v times beginbmatrix1 \ 0 \ 0endbmatrix$, the second column is $v times beginbmatrix0 \ 1 \ 0endbmatrix$,
and the third is $v times beginbmatrix0 \ 0 \ 1endbmatrix$.

If $B = e_1,dots,e_n$ and $D = f_1,dots,f_m$ and $T$ is a linear transformation, then $Phi_BD(T)$ is obtained by applying $T$ to each element of $B$ and witting the result in terms of $f_1,dots,f_m$. That is, if

$$ T(e_j) = sum_i=1^m a_i,jf_i, $$

then the $j$-th column of $Phi_BD(T)$ is

$$ beginbmatrix a_1,j \ a_2,j \ vdots \ a_m,j endbmatrix. $$

For example, $alpha_v(e_1) = v times e_1 = [0,a_3,-a_2]^T = 0e_1 + a_3e_2 -a_2e_3$ so the first column of $Phi_BB(alpha_v)$ is $[0,a_3,-a_2]^T$.