Transpose of a vector-vector product. I've this stupid question I cannot find the answer to.
Lets say I've a column vector $\mathbf v$. So now, the product $\mathbf{v}*\mathbf{v}^T$, being $\mathbf{v}^T$ the transpose of vector $\mathbf{v}$, will produce a square matrix $\mathbf{A}$.
So now, the transpose of matrix $\mathbf{A}$ will still be a square matrix, $\mathbf{A}^T$. I know this statement seems stupid, but keep reading.
Now I can say, "lets transpose the product of the vectors":
$\mathbf{A}^T=(\mathbf{v}*\mathbf{v}^T)^T$
But as you can distribute the transpose over the multiplication, you can say:
$(\mathbf{v}*\mathbf{v}^T)^T=\mathbf{v}^T*\mathbf{v}$
but the result of $\mathbf{v}^T*\mathbf{v}$ is a scalar.
So, for what I see:
$\mathbf{A}_{n,n}^T=(\mathbf{v}_{n,1}*\mathbf{v}_{1,n}^T)^T=\mathbf{v}_{1,n}^T*\mathbf{v}_{n,1}=K_{1,1}$
So if I do the operation in one way I obtain a scalar and if I solve if another way I obtain a matrix.
Yes, I know, there's something definitely wrong in my reasoning, but I cannot see where is the problem.
¿Where is my reasoning problem?
Thank you a lot.
 A: For two matrices $A$ and $B$, if the product $AB$ is well-defined, then $(AB)^T=B^TA^T$. You wrote $(vv^T)^T=v^Tv$, which is wrong, and should be corrected to $(vv^T)^T=(v^T)^Tv^T=vv^T$.
A: Let's try with  $\begin{bmatrix}3\\2\end{bmatrix}$ and its transposition $\begin{bmatrix}3&2\end{bmatrix}$. 
The product gives us :
$\begin{bmatrix}3\\2\end{bmatrix} * \begin{bmatrix}3&2\end{bmatrix} = \begin{bmatrix}9&6 \\ 6&4\end{bmatrix}^\intercal$
and
$\begin{bmatrix}9&6 \\ 6&4\end{bmatrix}^\intercal = \begin{bmatrix}9&6 \\ 6&4\end{bmatrix}$  
but (for you)
$\mathbf{A}^T=(\mathbf{v}*\mathbf{v}^T)^T$ and $(\mathbf{v}*\mathbf{v}^T)^T=\mathbf{v}^T*\mathbf{v}$
So  $\begin{bmatrix}9&6 \\ 6&4\end{bmatrix} =( \begin{bmatrix}3\\2\end{bmatrix} * \begin{bmatrix}3&2\end{bmatrix})^\intercal$ who is not equal to $\begin{bmatrix}3&2\end{bmatrix} * \begin{bmatrix}3\\2\end{bmatrix}$.
The reason is that $(\mathbf{A}*\mathbf{B})^T=\mathbf{B}^T*\mathbf{A}^T$
if $A =\mathbf{v}$ and $B =\mathbf{v}^T$  , $(\mathbf{B}^T*\mathbf{A}^T) = (\mathbf{v}*\mathbf{v}^T)$ and not $(\mathbf{v}^T*\mathbf{v})$
