From linear transformation to alternating linear transformation I'm reading Kenneth Hoffman's Linear Algebra, Ed2.
In $\S5.6$ "Multilinear Functions" it talks about generating an alternating linear transformation from a linear transformation.

The collection of all multilinear functions on $V$ will be denoted by
  $M^r(V)$. 
￼Definition. Let $L$ be an $r$-linear form on a $K$-module $V$. We say
  that $L$ is alternating if $L(\alpha_1, \dots, \alpha_r) = 0$
  whenever $\alpha_i = \alpha_j$ with $i \ne j$.  We denote by
  $\Lambda^r(V)$ the collection of all alternating $r$-linear forms on
  $V$.
There is a general method for associating an alternating form with a
  multilinear form. If $L$ is an $r$-linear form on a module $V$ and if
  $\sigma$ is a permutation of $\{1, . . . , r\}$, we obtain another
  $r$-linear function $L_\sigma$, by defining $$L\sigma(\alpha_1, \dots,
 \alpha_r) = L(\alpha_{\sigma 1}, \dots, \alpha_{\sigma_r}).$$ If $L$
  happens to be alternating, then $L\sigma = (\text{sgn }\sigma)L$. Now,
  for each $L$ in $M^r(V)$ we define a function $\pi_r L$ in $M^r( V)$
  by￼￼￼ ￼￼(5-35)  $$\pi_r L = \sum_\sigma(\text{sgn } \sigma) L_\sigma$$
  that is, (5-36) $$(\pi_r L) (\alpha_1, \dots, \alpha_r) =
\sum_\sigma(\text{sgn } \sigma) L(\alpha_{\sigma 1}, \dots,
 \alpha_{\sigma r}) $$
￼￼Lemma. $\pi_r$ is a linear transformation from $M^r(V)$ into
  $\Lambda^r(V)$. If $L$ is in $\Lambda^r(V)$ then $\pi_r L = r! L$.
￼Proof. Let $\tau$ be any permutation of $\{1, . . . , r\}$. Then 
  $$(\pi_r L)(\alpha_{\tau 1}, \dots, \alpha_{\tau r}) =
 \sum_\sigma(\text{sgn } \sigma) L(\alpha_{\tau \sigma 1}, \dots,
\alpha_{\tau \sigma r}) = (\text{sgn } \tau) \sum_\sigma(\text{sgn
 }\tau \sigma) L(\alpha_{\tau \sigma 1}, \dots, \alpha_{\tau \sigma
 r}).$$
As $\sigma$ runs (once) over all permutations of $\{1, . . . , r\}$,
  so does $\tau \sigma$. Therefore,  $$(\pi_r L)(\alpha_{\tau 1}, \dots,
 \alpha_{\tau r})  = (\text{sgn } \tau) (\pi_r L)    (\alpha_1, \dots,
 \alpha_r).$$
Thus $\pi_r L$ is an alternating form.
If $L$ is in $\Lambda^r(V)$, then $L(\alpha_{\sigma 1}, \dots,
 \alpha_{\sigma r}) = (\text{sgn } \sigma)L(\alpha_1, \dots, \alpha_r)$
  for each $\sigma$, hence $\pi_r L = r!L$.

These all look quite ok, but, I'm confused at an example:
Suppose I define $n=r=2$, and define $L$ as:
$$L(\alpha_1, \alpha_1) = 1, \quad L(\alpha_1, \alpha_2) = 2, \quad L(\alpha_2, \alpha_1) = 3, \quad L(\alpha_2, \alpha_2) = 4 $$
So $L$ could be interpreted by a matrix $\begin{pmatrix} 1 & 2 \\  3 & 4  \\  \end{pmatrix}$.
Then there are only two $\sigma$s: $\sigma_A = (1), \quad \sigma_B = (1,2)$.
 $L_{\sigma_A} = L$, and $L_{\sigma_B} = \begin{pmatrix} 4 & 3 \\  2 & 1  \\  \end{pmatrix}$
as
$L_{\sigma_B}(\alpha_1, \alpha_1) = L(\alpha_{\sigma_B 1}, \alpha_{\sigma_B 1}) = L(\alpha_2, \alpha_2) = 4$, $L_{\sigma_B}(\alpha_1, \alpha_2) = L(\alpha_{\sigma_B 1}, \alpha_{\sigma_B 2}) = L(\alpha_2, \alpha_1) = 3$, $L_{\sigma_B}(\alpha_2, \alpha_1) = L(\alpha_{\sigma_B 2}, \alpha_{\sigma_B 1}) = L(\alpha_1, \alpha_2) = 2$, 
$L_{\sigma_B}(\alpha_2, \alpha_2) = L(\alpha_{\sigma_B 2}, \alpha_{\sigma_B 2}) = L(\alpha_1, \alpha_1) = 1$.
Then $$\pi_2 L = \sum_\sigma (\text{sgn }\sigma) L_\sigma = (\text{sgn } \sigma_A) L_{\sigma_A} + (\text{sgn } \sigma_B) L_{\sigma_B} = L - L_{\sigma_B} = \begin{pmatrix} -3 & -1 \\ 1 & 3  \\  \end{pmatrix}.$$
However, if $\pi_2 L$ is alternating, by definition it requires $$(\pi_2L)(\alpha_1, \alpha_1) = 0.$$
The two conflict!
Now I'm lost here, why such two conflict, did I misunderstand on some topics?
 A: In the expression $L\sigma(\alpha_1,\ldots,\alpha_r) = L(\alpha_{\sigma1},\ldots,\alpha_{\sigma r})$, the $\alpha_1,\ldots,\alpha_r$ are denoting entires of the input and not specific basis vectors as you use them in your example. In particular, the right-hand side $L(\alpha_{\sigma1},\ldots,\alpha_{\sigma r})$ still uses the same inputs you originally had, only their order has been permuted. So, when $n=r=2$ you get
$$L_{\sigma_B}(\alpha_1,\alpha_2) = L_(\alpha_2,\alpha_1),$$
which means that to get the value of $L_{\sigma_B}$ on some pair of inputs, all you do is evaluate $L$ on the same inputs only with their order switched. In your example, the $(1,1)$-entry in the corresponding matrix for $L$ comes from evaluating $L$ on a pair $(v,v)$, where $v$ is your first basis vector (do NOT call this $\alpha_1$), so that in the notation above $\alpha_1=v$ and $\alpha_2=v$, and thus
$$L_{\sigma_B}(v,v) = L(v,v) = 1$$
is the $(1,1)$-entry in the matrix for $L_{\sigma_B}$. Similarly, the $(2,2)$-entry comes from evaluating at $(w,w)$ where $w$ (not $\alpha_2$) is your second basis vector, and switching the order of these two inputs still gives $(w,w)$.
