Finding the polar cone of the given cone Given a closed convex cone $D$ in $\mathbb{R}^{n}$, the cone $K_{2} \in \mathbb{R}^{m}$ is defined by $$ K_{2} = \{ y = (y^{1}, y^{2}, \cdots , y^{m}): y^{i} \in \mathbb{R}^{n},\, i= 1, \cdots , m, \, y^{1} + y^{2} + \cdots + y^{m} \in D \} $$
I need to describe its polar cone $K_{2}^{\circ}$.
Recall that for a given cone $C$, its polar cone $C^{\circ}$ is defined to be the set of all $x$ such that $\langle x,y \rangle \leq 0$ for all $y \in C$.
So, if $y \in K_{2}$, then $$ y = \begin{pmatrix} y^{1}, & y^{2},& \cdots, & y^{m}\end{pmatrix} \\ = \begin{pmatrix}\begin{pmatrix} y_{1}^{1} & y_{2}^{1} & \cdots y_{n}^{1} \end{pmatrix},  \begin{pmatrix} y_{1}^{2} & y_{2}^{2} & \cdots y_{n}^{2} \end{pmatrix},\cdots ,\begin{pmatrix} y_{1}^{m} & y_{2}^{m} & \cdots y_{n}^{m} \end{pmatrix} \end{pmatrix}.$$
So, I need to find the set of all $x$ such that when I take the inner product of $x$ and $y$, I get a value $\leq 0$.
My first problem is that I'm not sure if I have even expressed a general set in $K_{2}$ correctly here. Secondly, I would think that perhaps I should take the inner product of a general $y$ with a general $x$, set the result $\leq 0$ and then try to solve for what the components of $x$ are, but as I am not sure even what a general $x$ should look like, I am at a loss as to how this should be done.
If this is not the correct approach to finding the polar of this cone, what is the correct approach? Beyond the inner product definition of a polar cone (which in terms of angles between things, means that $x$ and $y$ make an obtuse angle with each other), I don't know much about how to go about finding them.
I sincerely thank you for your time and patience!
 A: Here's an initial observation, but not a full solution.  Changing notation slightly, 
$$K_2 = \{ Y = \begin{bmatrix} y_1 & \cdots & y_m \end{bmatrix} \in \mathbb R^{n \times m} \mid y_1 + \cdots + y_m \in D \}.$$ 
(Here $y_i$ is the $i$th column of the matrix $Y$.) A matrix $X \in \mathbb R^{m \times n}$ belongs to $K_2^\circ$ if and only if 
$$\langle X, Y \rangle = \text{tr}(Y X^T) = \text{tr}(y_1 x_1^T + \cdots + y_m x_m^T) \leq 0 \text{ for all } Y \in K_2.$$ If $x_1 = \cdots = x_m = x \in D^\circ$, then 
$$\langle X, Y \rangle = \text{tr}( (y_1+\cdots+y_m)^T x) = x^T(y_1 + \cdots + y_m) \leq 0,
$$ 
so $X \in K_2^\circ$.
This shows that $S = \{ \begin{bmatrix} x & \cdots & x \end{bmatrix} \mid x \in D^\circ\} \subset K_2^\circ$.
We can conjecture that in fact $S = K_2^\circ$.  We still need to show containment in the other direction.  
By the way, sometimes once you have guessed what the polar cone is, it turns out to be easier to show that $S^\circ = K_2$.  If it can be shown that $S^\circ = K_2$, which I suspect is straightforward, it will follow that $S = K_2^\circ$.

Let's attempt to show that $S^\circ \subset K_2$.  So, suppose that $Y \in S^\circ$.  From the definition of $S^\circ$, we have that $\langle Y, X \rangle \leq 0$ for all $X \in S$. Using the definition of $S$, we see that
$$
\tag{1}\langle Y, \begin{bmatrix} x & \cdots x \end{bmatrix} \rangle \leq 0
$$
for all $x \in \mathbb R^n$.
We have hoping to conclude that $Y \in K_2$.  In other words, we are hoping to conclude that the columns of $Y$ sum to $0$.  Does this follow somehow from the fact that the inequality (1) holds for all $x \in \mathbb R^n$?
A: The result follows from mostly formal cone calculations.
It is straightforward to show that if $K$ is a cone and $L$ a linear operator then
$(LK)^\circ = (L^T)^{-1} K^\circ$.
Let $A= \begin{bmatrix} I \cdots I \end{bmatrix}$, then $K_2 = A^{-1} D$. Note that this is the inverse in a set valued sense, $A$ is not
injective.
Note that this gives $A^{-1} D = \ker A + A^\dagger D $, where $A^\dagger$ is
the pseudo inverse of $A$.
Also note that $(\ker A)^\circ = {\cal R} A^T $ since $\ker A$ is
a subspace.
Note that $A^\dagger y = {1 \over n} A^Ty$, and so $A^\dagger D = A^T D$.
If $A,B$ are closed convex cones then $(A+B)^\circ = A^\circ \cap  B^\circ$.
Hence $K_2^\circ = (\ker A)^\circ \cap (A^T D)^\circ = {\cal R} A^T \cap (A^T D)^\circ = {\cal R} A^T \cap A^{-1} D^\circ$.
Noting that ${\cal R} A^T = \{ (y,\cdots, y) \}_{y \in \mathbb{R}^n}$,
we see that
$K_2^\circ = \{ (y,\cdots, y) | y \in D^\circ \} = A^T D^\circ$.
