Confused about transposes in kernel notation I am studying machine learning and I ran into a challenge that does not make sense to me. Maybe it is a crazy or a simple question.
If we have kernel $ \phi_1(x)=[x, x^2]^T $ and $ \phi_2(x)=[2x, 2x^2]^T $ and if $ \phi_2$ has greater margin  than  $ \phi_1$, can we say it is true for $ \phi_1(x)=[x, x^2] $and $ \phi_2(x)=[2x, 2x^2]$? Can we remove the transpose signs ($^T$)?
Here is a screenshot of a problem to provide some context:

 A: Taking a transpose doesn't really make sense. You define vectors as either rows or columns (based on the notation used here it appears column vectors are used) and then define the kernel/mapping function accordingly.
Basically, everything revolves around the following notation of inner product in feature space after using the mapping function $\phi(\cdot): \mathbb{R}^I \mapsto \mathbb{R}^F$ with $I$ and $F$ the input and feature space dimensionality, respectively (assuming column vectors):
$$\kappa(\mathbf{u}, \mathbf{v}) = \langle \phi(\mathbf{u}), \phi(\mathbf{v}) \rangle = \phi(\mathbf{u})^T \phi(\mathbf{v}).$$
The reason the correct answer is A is evident, as all distances using $\phi_2$ will be a factor 4 larger than those of $\phi_1$. $\phi_2$ is essentially the same mapping as $\phi_1$, but scaling each mapped coordinate by a factor 2. If you have two (column) vectors $\mathbf{u}$ and $\mathbf{v}$ and a feature-wise scaling factor $\alpha$, then:
$$\langle \mathbf{u}, \mathbf{v} \rangle = \mathbf{u}^T \mathbf{v}$$
$$\langle \alpha \mathbf{u}, \alpha \mathbf{v} \rangle = \alpha^2 \mathbf{u}^T \mathbf{v} = \alpha^2 \langle \mathbf{u}, \mathbf{v} \rangle$$
In the assignment $\phi_1(\mathbf{x}) = [\mathbf{x}, \mathbf{x}^2]^T$ and $\phi_2(\mathbf{x}) = [2\mathbf{x}, 2\mathbf{x}^2]^T = 2 \phi_1(\mathbf{x})$, which immediately means that all distances in the feature space induced by $\phi_2$ are a factor 4 larger than those in the space induced by $\phi_1$.
For the final notation you assume column vectors, but you might just as well use row vectors (then a dot product is $\langle \mathbf{u},\mathbf{v}\rangle=\mathbf{uv}^T$ instead of $\langle \mathbf{u},\mathbf{v}\rangle=\mathbf{u}^T\mathbf{v}$). It's a matter of definition. However, if you start working in column vectors you have to continue doing so.
