How to add vector with itself tranposed? So I'm solving basic linear algebra questions as part of review.
$$v=\begin{bmatrix} 1 & 2 & 3 \\ \end{bmatrix}$$
When I do the operation $v+v^T$ according to matlab, numpy and wolfram alpha
spits out
$$v+v^T=\begin{bmatrix} 2 & 3 & 4\\  3 & 4 & 5\\ 4 & 5 & 6\\ \end{bmatrix}$$
Originally I thought it would be like 
$$ v + v^T = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3\\ \end{bmatrix} + \begin{bmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 3 & 3\\ \end{bmatrix}=\begin{bmatrix} 2 & 3 & 4\\  3 & 4 & 5\\ 4 & 5 & 6\\ \end{bmatrix}$$
Can someone explain to be how this makes any sense?
 A: 
Can someone explain to be how this makes any sense?

It doesn't make any sense.
If you're talking about tansposition, then you're implicitly viewing your vectors as matrices with one of the dimensions being $1$.
Then your $v$ is a $1\times 3$ matrix and $v^T$ is a $3\times 1$ matrix.
Addition of matrices that don't have the same dimension is not defined.
A: Here is what Wolfram alpha is doing when you ask it for $v+v^T$ when $v$ is, say, a $3$-component vector.  If $v= (a, b, c)$ it will interpret $v+v^T$ as a sequence of three separate "shifts" of $v^T$:   first by $a$, then by $b$, then by $c$.  It puts these three shifts as columns of a matrix (in order), giving a $3 \times 3$ matrix as output.  So, if you shift $v^T$ by $a$ you get $(2a,a+b, a+c)^T$, and this will be the first column of the output matrix.  The second column is the shift by $b$ to get $(a+b, 2b, b+c)^T$, etc.  If you do this for your example, you get your claimed output above.  (I deduced this by following the step-by-step link in alpha where they show you what the computation $v+v^T$ is doing, so this is not speculation on my part.)
However, TO BE CLEAR, this is NOT traditional vector addition in any way whatsoever.   This is certainly undefined in your case.  (And all cases, except when your vectors have only one component, which is hardly a vector.)  It would be defined for square matrices of the same size. 
Moral:  never trust output unless you REALLY know what the machine is doing.  
A: A thought here as to why those systems would do that: there's an occasionally useful operation on a vector these systems support, namely adding a constant to each term. So then if $a$ is a constant and $v$ is a vector, the systems abbreviate this operator as $a+v$ or $v+a$. In effect, we promote $a$ to a vector with each element equal.
Now, if we're working with a row vector $u^T$ and a column vector $v$? In order to add the constants in each column of $u^T$ to $v$, we promote each of those length-1 columns to full length. In order to add the constants in each row of $v$ to $u^T$, we promote those length-1 rows to full length. This amounts to the same sum of matrices hypothesized in the question.
This is, of course, not standard vector/matrix addition.
