Suppose I have $m$ vectors $\mathbf{x_1}, \mathbf{x_2}, ... \mathbf{x_m} \in \mathbb{R}^n$. Each vector is divided into 3 parts, e.g. $\mathbf{x_i} = (\mathbf{y_i}, \mathbf{z_i}, \mathbf{w_i})$ where $\mathbf{y_i} \in \mathbb{R}^{n_1}$, $\mathbf{z_i} \in \mathbb{R}^{n_2}$, $\mathbf{w_i} \in \mathbb{R}^{n_3}$ and $n_1 + n_2 + n_3 = n$.
Consider the function $K(\mathbf{x_i}, \mathbf{x_j}) = \mathbf{y_i} \cdot \mathbf{y_j} + \mathbf{z_i} \cdot \mathbf{z_j} + \mathbf{w_i} \cdot \mathbf{w_j}$. It can be verified that $K$ satisfies the 3 axioms of an inner product (in fact, $K(\mathbf{x_i}, \mathbf{x_j}) = \mathbf{x_i} \cdot \mathbf{x_j}$).
However, suppose $K$ is instead defined as $K(\mathbf{x_i}, \mathbf{x_j}) = (\mathbf{y_i} \cdot \mathbf{y_j})(\mathbf{z_i} \cdot \mathbf{z_j} + \mathbf{w_i} \cdot \mathbf{w_j})$. $K$ still satisfies the symmetry requirement, but no longer satisfies axioms (2) and (3).
Here is my question: does there exist a mapping $\phi$ under which $K$ corresponds to an inner product, e.g. $K(\mathbf{x_i}, \mathbf{x_j}) = \langle \phi(\mathbf{x_i}), \phi(\mathbf{x_j}) \rangle$? If not, can you prove otherwise?
Define \begin{equation} K(\mathbf{x_i}, \mathbf{x_j}) = (\mathbf{y_i} \cdot \mathbf{y_j})(\mathbf{z_i} \cdot \mathbf{z_j} + \mathbf{w_i} \cdot \mathbf{w_j}) \end{equation} Suppose there exists a mapping $\phi$ satisfying \begin{equation} K(\mathbf{x_i}, \mathbf{x_j}) = \langle \phi(\mathbf{x_i}), \phi(\mathbf{x_j}) \rangle \end{equation} A desirable property for such a $\phi$ would be that for $\alpha\in\mathbb{R}$ \begin{equation} \phi(\alpha\mathbf{x})=\alpha^2\phi(\mathbf{x}) \end{equation} since for $\alpha\in\mathbb{R}$ we have \begin{equation} K(\alpha\mathbf{x_i}, \mathbf{x_j})=\langle \phi(\alpha\mathbf{x_i}), \phi(\mathbf{x_j}) \rangle \end{equation} Also, it is the case that \begin{eqnarray} K(\alpha\mathbf{x_i}, \mathbf{x_j}) &=& (\alpha\mathbf{y_i} \cdot \mathbf{y_j})(\alpha\mathbf{z_i} \cdot \mathbf{z_j} + \alpha\mathbf{w_i} \cdot \mathbf{w_j})\\ &=&\alpha^2 (\mathbf{y_i} \cdot \mathbf{y_j})(\mathbf{z_i} \cdot \mathbf{z_j} + \mathbf{w_i} \cdot \mathbf{w_j})\\ &=&\alpha^2 \langle \phi(\mathbf{x_i}), \phi(\mathbf{x_j}) \rangle \end{eqnarray} Therefore \begin{eqnarray} \langle \phi(\alpha\mathbf{x_i}), \phi(\mathbf{x_j}) \rangle&=&\alpha^2 \langle \phi(\mathbf{x_i}), \phi(\mathbf{x_j}) \rangle\\ &=&\langle \alpha^2\phi(\mathbf{x_i}), \phi(\mathbf{x_j}) \rangle \end{eqnarray} This would follow provided \begin{equation} \phi(\alpha\mathbf{x_i})=\alpha^2\phi(\mathbf{x_i}) \end{equation} This leaves open the question of whether such a $\phi$ exists.
Note that whereas $$ \langle \phi(\alpha\mathbf{x_i}), \phi(\mathbf{x_j}) \rangle=\langle \alpha^2\phi(\mathbf{x_i}), \phi(\mathbf{x_j}) \rangle $$ does not imply that $$ \phi(\alpha\mathbf{x_i})=\alpha^2\phi(\mathbf{x_i}) $$ it does imply that
$$ \vert \phi(\alpha\mathbf{x})\vert=\alpha^2\vert\phi(\mathbf{x})\vert $$
if $\phi$ is real-valued, since
\begin{eqnarray} \vert\phi(\alpha\mathbf{x})\vert^2&=&\langle\phi(\alpha\mathbf{x}),\phi(\alpha\mathbf{x})\rangle\\ &=&\langle\alpha^2\phi(\mathbf{x}),\phi(\alpha\mathbf{x})\rangle\\ &=&\alpha^2\langle\phi(\alpha\mathbf{x}),\phi(\mathbf{x})\rangle\\ &=&\alpha^2\langle\alpha^2\phi(\mathbf{x}),\phi(\mathbf{x})\rangle\\ &=&\alpha^4\langle\phi(\mathbf{x}),\phi(\mathbf{x})\rangle\\ &=&\alpha^4\vert\phi(\mathbf{x})\vert^2 \end{eqnarray}