A typographical error let to an unexpected (but, for me, potentially useful) result:
$$ \left\{\begin{array} & a & b & c\\ d & e & f \\ g & h & i \end{array}\right\}\left\{1, 2, 3\right\} $$
Yielding the result $$ \left\{\begin{array} 1a & 1b & 1c\\ 2d & 2e & 2f \\ 3g & 3h & 3i \end{array}\right\} $$
Unfortunately, with nothing to google I can't find out what the operation is called, and would appreciate any assistance.
This is the typographical error and the resultant math at wolframalpha
When you multiply a matrix by a scalar, every entry in the matrix gets multiplied by that amount. E.g.: $2\cdot \left[\begin{smallmatrix} a&b\\c&d\end{smallmatrix}\right]=\left[\begin{smallmatrix} 2a & 2b\\ 2c&2d\end{smallmatrix}\right]$
Wolfram seems to have interpreted your input as something similar, which may have precedence in coding, but does not (to my knowledge) have its own name or notation. It appears to have taken the vector as a list of scalars to multiply by, and applied the multiplication to each respective row separately. $\{1,2\}\cdot \left[\begin{smallmatrix} a&b\\c&d\end{smallmatrix}\right] = \left[\begin{smallmatrix}a&b\\2c&2d\end{smallmatrix}\right]$. (here I use { } instead of [ ] for the "list" to emphasize my expectation that wolfram is not treating this as a matrix)
If I were to coin a term for this, I might call it "row-wise scalar multiplication." There doesn't seem to be any canonical uses for this, and it does not appear in anything I have personally read. We can define the operation as, $l \cdot A = A\cdot l = (l_i\cdot a_{i,j})_{i,j}$ where $l$ is a list of length $m$, and $A$ is an $m\times n$ matrix.
If you prefer to use established operations, by rewording the length $m$ list, $l = \{l_1, l_2,\dots, l_m\}$, as an $m\times n$ matrix, $L=\left[\begin{smallmatrix}l_1&\dots& l_1\\\vdots & &\vdots\\l_m&\cdots & l_m\end{smallmatrix}\right]$, the previously defined operation can be worded instead simply using the hadamard product as $L\circ A = \left[\begin{smallmatrix} l_1\cdot a_{1,1}&\cdots & l_1\cdot a_{1,n}\\ \vdots & &\vdots\\ l_m\cdot a_{m,1}&\cdots & l_m\cdot a_{m,n}\end{smallmatrix}\right]$. We had to change what objects we are using in order to have the dimensions match correctly, but this shouldn't provide any great hurdle in terms of programming code or writing proofs.
This form is more useful to notice certain properties, such as the fact that since hadamard multiplication is commutative, the property you observed in your later post is trivially true $(l^{\frac{1}{2}}\cdot A)\cdot l^{\frac{1}{2}} = l\cdot A$ (where exponents on a list are applied entry-wise).