If I have a matrix $V$ which has vector columns $v_1,v_2,\ldots,v_d$, and suppose that I have a function that maps $$\phi(x)=(v_1^\top x,\ldots, v_k^\top x)^\top$$
Now, if $V_k=[v_1\,\, v_2\,\, \ldots\,\,v_k]$, then how can I show that $$\phi(x_i)^\top\phi(x_j)=x_i^\top V_kV_k^\top x_j$$
My Thoughts: So, I understand that $\phi$ is essentially projecting a dimension $d$ vector onto the linear span of a set $\{v_1,\ldots,v_k\}$. But, I’m not sure how to proceed from here.
You defined $\phi(x)$ as $$\phi(x) = \begin{bmatrix}v_1^Tx \\ \vdots \\ v_k^Tx\end{bmatrix}$$ The vectors $v_i^T$ are the row vectors of $V_k^T$, so $\phi(x)$ is given by $$\phi(x) = V_k^Tx$$ Then $\phi(x)^T$ is $$\phi(x)^T=(V_k^Tx)^T = x^TV_k$$ As a result, $\phi(x_i)^T\phi(x_j)$ may be written $$\phi(x_i)^T\phi(x_j) = x_i^TV_kV_k^Tx_j$$