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?

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.  
