So far, the majority of the machine learning models we have seen have been linear.
In this lecture, we will see a general way to make many of these models non-linear. We willl use a new idea called kernels.
Recall that a linear model has the form $$ f(x) = \sum_{j=0}^d \theta_j \cdot x_j = \theta^\top x. $$ where $x$ is a vector of features and we used the notation $x_0 = 1$.
We pick $\theta$ to minimize the (L2-regularized) mean squared error (MSE): $$J(\theta)= \frac{1}{2n} \sum_{i=1}^n(y^{(i)} - \theta^\top x^{(i)})^2 + \frac{\lambda}{2}\sum_{j=1}^d \theta_j^2$$
Recall that a polynomial of degree $p$ is a function of the form $$ a_p x^p + a_{p-1} x^{p-1} + ... + a_{1} x + a_0. $$
Below are some examples of polynomial functions.
Specifically, given a one-dimensional continuous variable $x$, we can defining a feature function $\phi : \mathbb{R} \to \mathbb{R}^{p+1}$ as $$ \phi(x) = \begin{bmatrix} 1 \\ x \\ x^2 \\ \vdots \\ x^p \end{bmatrix}. $$
The class of models of the form $$ f_\theta(x) := \sum_{j=0}^p \theta_p x^p = \theta^\top \phi(x) $$ with parameters $\theta$ and polynomial features $\phi$ is the set of $p$-degree polynomials.
Any non-linear feature map $\phi(x) : \mathbb{R}^d \to \mathbb{R}^p$ can be used to obtain general models of the form $$ f_\theta(x) := \theta^\top \phi(x) $$ that are highly non-linear in $x$ but linear in $\theta$.
It is useful to represent the featurized dataset as a matrix $\Phi \in \mathbb{R}^{n \times p}$:
$$ \Phi = \begin{bmatrix} \phi(x^{(1)})_1 & \phi(x^{(1)})_2 & \ldots & \phi(x^{(1)})_p \\ \phi(x^{(2)})_1 & \phi(x^{(2)})_2 & \ldots & \phi(x^{(2)})_p \\ \vdots \\ \phi(x^{(n)})_1 & \phi(x^{(n)})_2 & \ldots & \phi(x^{(n)})_p \end{bmatrix} = \begin{bmatrix} - & \phi(x^{(1)})^\top & - \\ - & \phi(x^{(2)})^\top & - \\ & \vdots & \\ - & \phi(x^{(n)})^\top & - \\ \end{bmatrix} .$$The normal equations provide a closed-form solution for $\theta$: $$ \theta = (X^\top X + \lambda I)^{-1} X^\top y.$$
When the vectors of attributes $x^{(i)}$ are featurized, we can write this as $$ \theta = (\Phi^\top \Phi + \lambda I)^{-1} \Phi^\top y.$$
We can modify this expression by using a version of the push-through matrix identity: $$ (\lambda I + U V)^{-1} U = U (\lambda I + V U)^{-1} $$ where $U \in \mathbb{R}^{n \times m}$ and $V \in \mathbb{R}^{m \times n}$ and $\lambda \neq 0$
Proof sketch: Start with $U (\lambda I + V U) = (\lambda I + U V) U$ and multiply both sides by $(\lambda I + V U)^{-1}$ on the right and $(\lambda I + U V)^{-1}$ on the left.
We can apply the identity $(\lambda I + U V)^{-1} U = U (\lambda I + V U)^{-1}$ to the normal equations with $U=\Phi^\top$ and $V=\Phi$.
$$ \theta = (\Phi^\top \Phi + \lambda I)^{-1} \Phi^\top y$$to obtain the dual form:
$$ \theta = \Phi^\top (\Phi \Phi^\top + \lambda I)^{-1} y.$$The first approach takes $O(p^3)$ time; the second is $O(n^3)$ and is faster when $p > n$.
An interesting corollary of the dual form $$ \theta = \Phi^\top \underbrace{(\Phi \Phi^\top + \lambda I)^{-1} y}_\alpha$$ is that the optimal $\theta$ is a linear combination of the $n$ training set features: $$ \theta = \sum_{i=1}^n \alpha_i \phi(x^{(i)}). $$
Here, the weights $\alpha_i$ are derived from $(\Phi \Phi^\top + \lambda I)^{-1} y$ and equal $$\alpha_i = \sum_{j=1}^n L_{ij} y_j$$ where $L = (\Phi \Phi^\top + \lambda I)^{-1}.$
Consider now a prediction $\phi(x')^\top \theta$ at a new input $x'$: $$\phi(x')^\top \theta = \sum_{i=1}^n \alpha_i \phi(x')^\top \phi(x^{(i)}).$$
The crucial observation is that the features $\phi(x)$ are never used directly in this equation. Only their dot product is used!
This observation will be at the heart of a powerful new idea called the kernel trick.
We also don't need features $\phi$ for learning $\theta$, just their dot product! First, recall that each row $i$ of $\Phi$ is the $i$-th featurized input $\phi(x^{(i)})^\top$.
Thus $K = \Phi \Phi^\top$ is a matrix of all dot products between all the $\phi(x^{(i)})$ $$K_{ij} = \phi(x^{(i)})^\top \phi(x^{(j)}).$$
We can compute $\alpha = (K+\lambda I)^{-1}y$ and use it for predictions $$\phi(x')^\top \theta = \sum_{i=1}^n \alpha_i \phi(x')^\top \phi(x^{(i)}).$$ and all this only requires dot products, not features $\phi$!
The above observations hint at a powerful new idea -- if we can compute dot products of features $\phi(x)$ efficiently, then we will be able to use high-dimensional features easily.
It turns our that we can do this for many ML algorithms -- we call this the Kernel Trick.
Many ML algorithms can be written down as optimization problems in which the features $\phi(x)$ only appear as dot products $\phi(x)^\top \phi(z)$ that can be computed efficiently.
Let's look at an example.
Recall that a linear model has the form $$ f(x) = \sum_{j=0}^d \theta_j \cdot x_j = \theta^\top x. $$ where $x$ is a vector of features and we used the notation $x_0 = 1$.
Any non-linear feature map $\phi(x) : \mathbb{R}^d \to \mathbb{R}^p$ can be used in this way to obtain general models of the form $$ f_\theta(x) := \theta^\top \phi(x) $$ that are highly non-linear in $x$ but linear in $\theta$.
It is useful to represent the featurized dataset as a matrix $\Phi \in \mathbb{R}^{n \times p}$:
$$ \Phi = \begin{bmatrix} \phi(x^{(1)})_1 & \phi(x^{(1)})_2 & \ldots & \phi(x^{(1)})_p \\ \phi(x^{(2)})_1 & \phi(x^{(2)})_2 & \ldots & \phi(x^{(2)})_p \\ \vdots \\ \phi(x^{(n)})_1 & \phi(x^{(n)})_2 & \ldots & \phi(x^{(n)})_p \end{bmatrix} = \begin{bmatrix} - & \phi(x^{(1)})^\top & - \\ - & \phi(x^{(2)})^\top & - \\ & \vdots & \\ - & \phi(x^{(n)})^\top & - \\ \end{bmatrix} .$$The normal equations provide a closed-form solution for $\theta$:
$$ \theta = (\Phi^\top \Phi + \lambda I)^{-1} \Phi^\top y.$$They also can be written in this form:
$$ \theta = \Phi^\top (\Phi \Phi^\top + \lambda I)^{-1} y.$$The first approach takes $O(d^3)$ time; the second is $O(n^3)$ and is faster when $d > n$.
An interesting corollary is that the optimal $\theta$ is a linear combination of the $n$ training set features: $$ \theta = \sum_{i=1}^n \alpha_i \phi(x^{(i)}). $$ We can compute a prediction $\phi(x')^\top \theta$ for $x'$ without ever using the features (only their dot products): $$\phi(x')^\top \theta = \sum_{i=1}^n \alpha_i \phi(x')^\top \phi(x^{(i)}).$$ Equally importantly, we can learn $\theta$ from only dot products.
Note that a $p$-th degree polynomial
$$ a_p x^p + a_{p-1} x^{p-1} + ... + a_{1} x + a_0. $$forms a linear model with parameters $a_p, a_{p-1}, ..., a_0$. This means we can use our algorithms for linear models to learn non-linear features!
Specifically, given a one-dimensional continuous variable $x$, we can defining a feature function $\phi : \mathbb{R} \to \mathbb{R}^p$ as
$$\phi(x) = \begin{bmatrix} 1 \\ x \\ x^2 \\ \vdots \\ x^p \end{bmatrix}. $$Then the class of models of the form $$ f_\theta(x) := \sum_{j=0}^p \theta_p x^p = \theta^\top \phi(x) $$ with parameters $\theta$ encompasses the set of $p$-degree polynomials. Specifically,
Can we compute the dot product $\phi(x)^\top \phi(x')$ of polynomial features $\phi(x)$ more efficiently than using the standard definition of a dot product? Let's look at an example.
To start, consider polynomial features $\phi : \mathbb{R}^d \to \mathbb{R}^{d^2}$ of the form
$$ \phi(x)_{ij} = x_i x_j \;\text{ for $i,j \in \{1,2,\ldots,d\}$}. $$For $d=3$ this looks like $$ \small \phi(x) = \begin{bmatrix} x_1 x_1 \\ x_1 x_2 \\ x_1 x_3 \\ x_2 x_1 \\ x_2 x_1 \\ x_2 x_2 \\ x_3 x_3 \\ x_3 x_1 \\ x_3 x_2 \\ x_3 x_3 \\ \end{bmatrix}. $$
The product of $x$ and $z$ in feature space equals: $$ \phi(x)^\top \phi(z) = \sum_{i=1}^d \sum_{j=1}^d x_i x_j z_i z_j $$ Computing this dot product invovles the sum over $d^2$ terms and takes $O(d^2)$ time.
An altenative way of computing the dot product $\phi(x)^\top \phi(z)$ is to instead compute $(x^\top z)^2$. One can check that this has the same result: \begin{align*} (x^\top z)^2 & = (\sum_{i=1}^d x_i z_i)^2 \\ & = (\sum_{i=1}^d x_i z_i) \cdot (\sum_{j=1}^d x_j z_j) \\ & = \sum_{i=1}^d \sum_{j=1}^d x_i z_i x_j z_j \\ & = \phi(x)^\top \phi(z) \end{align*}
However, computing $(x^\top z)^2$ can be done in only $O(d)$ time!
This is a very powerful idea:
The number of polynomial features $\phi_p$ of degree $p$ when $x \in \mathbb{R}^d$
$$ \phi_p(x)_{i_1, i_2, \ldots, i_p} = x_{i_1} x_{i_2} \cdots x_{i_p} \;\text{ for $i_1, i_2, \ldots, i_p \in \{1,2,\ldots,d\}$} $$scales as $O(d^p)$.
However, we can compute the dot product $\phi_p(x)^\top \phi_p(z)$ in this feature space in only $O(d)$ time for any $p$ as: $$\phi_p(x)^\top \phi_p(z) = (x^\top z)^p.$$
Many types of features $\phi(x)$ have the property that their dot product $\phi(x)^\top \phi(z)$ can be computed more efficiently than if we had to form these features explicitly.
Also, we will see that many algorithms in machine learning can be written down as optimization problems in which the features $\phi(x)$ only appear as dot products $\phi(x)^\top \phi(z)$.
The Kernel Trick means that we can use complex non-linear features within these algorithms with little additional computational cost.
Examples of algorithms in which we can use the Kernel trick:
We will look at more examples shortly.
Many ML algorithms can be written down as optimization problems in which the features $\phi(x)$ only appear as dot products $\phi(x)^\top \phi(z)$ that can be computed efficiently.
We will now see how SVMs can benefit from the Kernel Trick as well.
Consider a training dataset $\mathcal{D} = \{(x^{(1)}, y^{(1)}), (x^{(2)}, y^{(2)}), \ldots, (x^{(n)}, y^{(n)})\}$.
We distinguish between two types of supervised learning problems depnding on the targets $y^{(i)}$.
In this lecture, we assume $\mathcal{Y} = \{-1, +1\}$.
We will consider models of the form
\begin{align*} f_\theta(x) = \theta^\top \phi(x) + \theta_0 \end{align*}where $x$ is the input and $y \in \{-1, 1\}$ is the target.
Recall that the the max-margin hyperplane can be formualted as the solution to the following primal optimization problem. \begin{align*} \min_{\theta,\theta_0, \xi}\; & \frac{1}{2}||\theta||^2 + C \sum_{i=1}^n \xi_i \; \\ \text{subject to } \; & y^{(i)}((x^{(i)})^\top\theta+\theta_0)\geq 1 - \xi_i \; \text{for all $i$} \\ & \xi_i \geq 0 \end{align*}
The solution to this problem also happens to be given by the following dual problem: \begin{align*} \max_{\lambda} & \sum_{i=1}^n \lambda_i - \frac{1}{2} \sum_{i=1}^n \sum_{k=1}^n \lambda_i \lambda_k y^{(i)} y^{(k)} (x^{(i)})^\top x^{(k)} \\ \text{subject to } \; & \sum_{i=1}^n \lambda_i y^{(i)} = 0 \\ & C \geq \lambda_i \geq 0 \; \text{for all $i$} \end{align*}
We can obtain a primal solution from the dual via the following equation: $$ \theta^* = \sum_{i=1}^n \lambda_i^* y^{(i)} \phi(x^{(i)}). $$
Ignoring the $\theta_0$ term for now, the score at a new point $x'$ will equal $$ \theta^\top \phi(x') = \sum_{i=1}^n \lambda_i^* y^{(i)} \phi(x^{(i)})^\top \phi(x'). $$
Notice that in both equations, the features $x$ are never used directly. Only their dot product is used. \begin{align*} \sum_{i=1}^n \lambda_i - \frac{1}{2} \sum_{i=1}^n \sum_{k=1}^n \lambda_i \lambda_k y^{(i)} y^{(k)} \phi(x^{(i)})^\top \phi(x^{(k)}) \\ \theta^\top \phi(x') = \sum_{i=1}^n \lambda_i^* y^{(i)} \phi(x^{(i)})^\top \phi(x'). \end{align*}
If we can compute the dot product efficiently, we can potentially use very complex features.
More generally, given features $\phi(x)$, suppose that we have a function $K : \mathcal{X} \times \mathcal{X} \to [0, \infty]$ that outputs dot products between vectors in $\mathcal{X}$
$$ K(x, z) = \phi(x)^\top \phi(z). $$We will call $K$ the kernel function.
Recall that an example of a useful kernel function is $$K(x,z) = (x \cdot z)^p$$ because it computes the dot product of polynomial features of degree $p$.
Then notice that we can rewrite the dual of the SVM as \begin{align*} \max_{\lambda} & \sum_{i=1}^n \lambda_i - \frac{1}{2} \sum_{i=1}^n \sum_{k=1}^n \lambda_i \lambda_k y^{(i)} y^{(k)} K(x^{(i)}, x^{(k)}) \\ \text{subject to } \; & \sum_{i=1}^n \lambda_i y^{(i)} = 0 \\ & C \geq \lambda_i \geq 0 \; \text{for all $i$} \end{align*} and predictions at a new point $x'$ are given by $\sum_{i=1}^n \lambda_i^* y^{(i)} K(x^{(i)}, x').$
Using our earlier trick, we can use polynomial features of any degree $p$ in SVMs without forming these features and at no extra cost!
Now that we saw the kernel trick, let's look at several examples of kernels.
We will consider models of the form
\begin{align*} f_\theta(x) = \theta^\top \phi(x) + \theta_0 \end{align*}where $x$ is the input and $y$ is the target.
The normal equations provide a closed-form solution for $\theta$:
$$ \theta = (\Phi^\top \Phi + \lambda I)^{-1} \Phi^\top y.$$They also can be written in this form:
$$ \theta = \Phi^\top (\Phi \Phi^\top + \lambda I)^{-1} y.$$The first approach takes $O(d^3)$ time; the second is $O(n^3)$ and is faster when $d > n$.
An interesting corollary is that the optimal $\theta$ is a linear combination of the $n$ training set features: $$ \theta = \sum_{i=1}^n \alpha_i \phi(x^{(i)}). $$ We can compute a prediction $\phi(x')^\top \theta$ for $x'$ without ever using the features (only their dot products): $$\phi(x')^\top \theta = \sum_{i=1}^n \alpha_i \phi(x')^\top \phi(x^{(i)}).$$ Equally importantly, we can learn $\theta$ from only dot products.
Notice that in both equations, the features $x$ are never used directly. Only their dot product is used. \begin{align*} \sum_{i=1}^n \lambda_i - \frac{1}{2} \sum_{i=1}^n \sum_{k=1}^n \lambda_i \lambda_k y^{(i)} y^{(k)} \phi(x^{(i)})^\top \phi(x^{(k)}) \\ \theta^\top \phi(x') = \sum_{i=1}^n \lambda_i^* y^{(i)} \phi(x^{(i)})^\top \phi(x'). \end{align*}
If we can compute the dot product efficiently, we can potentially use very complex features.
The kernel corresponding to features $\phi(x)$ is a function $K : \mathcal{X} \times \mathcal{X} \to [0, \infty]$ that outputs dot products between vectors in $\mathcal{X}$ $$ K(x, z) = \phi(x)^\top \phi(z). $$
We will also consider general functions $K : \mathcal{X} \times \mathcal{X} \to [0, \infty]$ and call these kernel functions.
Kernels have various intepreations:
In order to illustrate kernels, we will use this dataset.
# https://scikit-learn.org/stable/auto_examples/svm/plot_svm_kernels.html
import numpy as np
import matplotlib.pyplot as plt
from sklearn import svm
# Our dataset and targets
X = np.c_[(.4, -.7), (-1.5, -1), (-1.4, -.9), (-1.3, -1.2), (-1.1, -.2), (-1.2, -.4), (-.5, 1.2), (-1.5, 2.1), (1, 1),
(1.3, .8), (1.2, .5), (.2, -2), (.5, -2.4), (.2, -2.3), (0, -2.7), (1.3, 2.1)].T
Y = [0] * 8 + [1] * 8
x_min, x_max = -3, 3
y_min, y_max = -3, 3
plt.scatter(X[:, 0], X[:, 1], c=Y, zorder=10, cmap=plt.cm.Paired, edgecolors='k', s=80)
plt.xlim(x_min, x_max)
plt.ylim(y_min, y_max)
(-3.0, 3.0)
The simplest kind of kernel that exists is called the linear kernel. This simply corresponds to dot product multiplication of the features: $$K(x,z) = x^\top z$$
Applied to an SVM, this corresponds to a linear decision boundary.
Below is an example of how we can use the SVM implementation in sklearn
with a linear kernel.
Internally, this solves the dual SVM optimization problem.
# https://scikit-learn.org/stable/auto_examples/svm/plot_svm_kernels.html
clf = svm.SVC(kernel='linear', gamma=2)
clf.fit(X, Y)
# plot the line, the points, and the nearest vectors to the plane
plt.scatter(clf.support_vectors_[:, 0], clf.support_vectors_[:, 1], s=80, facecolors='none', zorder=10, edgecolors='k')
plt.scatter(X[:, 0], X[:, 1], c=Y, zorder=10, cmap=plt.cm.Paired, edgecolors='k')
XX, YY = np.mgrid[x_min:x_max:200j, y_min:y_max:200j]
Z = clf.decision_function(np.c_[XX.ravel(), YY.ravel()])
# Put the result into a color plot
Z = Z.reshape(XX.shape)
plt.pcolormesh(XX, YY, Z > 0, cmap=plt.cm.Paired)
plt.contour(XX, YY, Z, colors=['k', 'k', 'k'], linestyles=['--', '-', '--'],levels=[-.5, 0, .5])
plt.xlim(x_min, x_max)
plt.ylim(y_min, y_max)
(-3.0, 3.0)
A more interesting example is the polynomial kernel of degree $p$, of which we have already seen a simple example: $$K(x,z) = (x^\top z + c)^p.$$
This corresponds to a mapping to a feature space of dimension $d+p \choose p$ that has all monomials $x_{i_1}x_{i_2}\cdots x_{i_p}$ of degree at most $p$.
For $d=3$ this feature map looks like $$ \small \phi(x) = \begin{bmatrix} x_1 x_1 \\ x_1 x_2 \\ x_1 x_3 \\ x_2 x_1 \\ x_2 x_1 \\ x_2 x_2 \\ x_3 x_3 \\ x_3 x_1 \\ x_3 x_2 \\ x_3 x_3 \\ \sqrt{2}c x_1 \\ \sqrt{2}c x_2 \\ \sqrt{2}c x_3 \\ c \end{bmatrix}. $$
The polynomial kernel allows us to compute dot products in a $O(d^p)$-dimensional space in time $O(d)$.
Let's see how it would be implemented in sklearn
.
# https://scikit-learn.org/stable/auto_examples/svm/plot_svm_kernels.html
clf = svm.SVC(kernel='poly', degree=3, gamma=2)
clf.fit(X, Y)
# plot the line, the points, and the nearest vectors to the plane
plt.scatter(clf.support_vectors_[:, 0], clf.support_vectors_[:, 1], s=80, facecolors='none', zorder=10, edgecolors='k')
plt.scatter(X[:, 0], X[:, 1], c=Y, zorder=10, cmap=plt.cm.Paired, edgecolors='k')
XX, YY = np.mgrid[x_min:x_max:200j, y_min:y_max:200j]
Z = clf.decision_function(np.c_[XX.ravel(), YY.ravel()])
# Put the result into a color plot
Z = Z.reshape(XX.shape)
plt.pcolormesh(XX, YY, Z > 0, cmap=plt.cm.Paired)
plt.contour(XX, YY, Z, colors=['k', 'k', 'k'], linestyles=['--', '-', '--'],levels=[-.5, 0, .5])
plt.xlim(x_min, x_max)
plt.ylim(y_min, y_max)
(-3.0, 3.0)
Another example is the Radial Basis Function (RBF; sometimes called Gaussian) kernel $$K(x,z) = \exp \left(-\frac{||x - z||^2}{2\sigma^2}\right),$$ where $\sigma$ is a hyper-parameter. It's easiest to understand this kernel by viewing it as a similarity measure.
We can show that this kernel corresponds to an infinite-dimensional feature map and the limit of the polynomial kernel as $p \to \infty$.
To see why that's intuitively the case, consider the Taylor expansion
$$ \exp \left(-\frac{||x - z||^2}{2\sigma^2}\right) \approx 1 - \frac{||x - z||^2}{2\sigma^2} + \frac{||x - z||^4}{2! \cdot 4\sigma^4} - \frac{||x - z||^6}{3! \cdot 8\sigma^6} + \ldots $$Each term on the right hand side can be expanded into a polynomial.
We can look at the sklearn
implementation again.
# https://scikit-learn.org/stable/auto_examples/svm/plot_svm_kernels.html
clf = svm.SVC(kernel='rbf', gamma=.5)
clf.fit(X, Y)
# plot the line, the points, and the nearest vectors to the plane
plt.scatter(clf.support_vectors_[:, 0], clf.support_vectors_[:, 1], s=80, facecolors='none', zorder=10, edgecolors='k')
plt.scatter(X[:, 0], X[:, 1], c=Y, zorder=10, cmap=plt.cm.Paired, edgecolors='k')
XX, YY = np.mgrid[x_min:x_max:200j, y_min:y_max:200j]
Z = clf.decision_function(np.c_[XX.ravel(), YY.ravel()])
# Put the result into a color plot
Z = Z.reshape(XX.shape)
plt.pcolormesh(XX, YY, Z > 0, cmap=plt.cm.Paired)
plt.contour(XX, YY, Z, colors=['k', 'k', 'k'], linestyles=['--', '-', '--'],levels=[-.5, 0, .5])
plt.xlim(x_min, x_max)
plt.ylim(y_min, y_max)
(-3.0, 3.0)
We've seen that for many features $\phi$ we can define a kernel function $K : \mathcal{X} \times \mathcal{X} \to [0, \infty]$ that efficiently computes $\phi(x)^\top \phi(x)$.
Suppose now that we use some kernel function $K : \mathcal{X} \times \mathcal{X} \to [0, \infty]$ in an ML algorithm. Is there an implicit feature mapping $\phi$ that corresponds to using K?
Let's start by defining a necessary condition for $K : \mathcal{X} \times \mathcal{X} \to [0, \infty]$ to be associated with a feature map.
Suppose that $K$ is a kernel for some feature map $\phi$, and consider an arbitrary set of $n$ points $\{x^{(1)}, x^{(2)}, \ldots, x^{(n)}\}$.
Consider the matrix $L \in \mathbb{R}^{n\times n}$ defined as $L_{ij} = K(x^{(i)}, x^{(j)}) = \phi(x^{(i)})^\top \phi(x^{(j)})$. We claim that $L$ must be symmetric and positive semidefinite.
Indeed, it $L$ is symmetric because the dot product $\phi(x^{(i)})^\top \phi(x^{(j)})$ is symmetric. Moreover, for any $z$,
\begin{align*} z^\top L z & = \sum_{i=1}^n \sum_{j=1}^n z_i L_{ij} z_j = \sum_{i=1}^n \sum_{j=1}^n z_i \phi(x^{(i)})^\top \phi(x^{(j)}) z_j \\ & = \sum_{i=1}^n \sum_{j=1}^n z_i (\sum_{k=1}^n \phi(x^{(i)})_k \phi(x^{(j)})_k ) z_j \\ & = \sum_{k=1}^n \sum_{i=1}^n \sum_{j=1}^n z_i \phi(x^{(i)})_k \phi(x^{(j)})_k z_j \\ & = \sum_{k=1}^n \sum_{i=1}^n \left( z_i \phi(x^{(i)})_k \right)^2 \geq 0 \end{align*}Thus if $K$ is a kernel, $L$ must be positive semidefinite for any $n$ points $x^{(i)}$.
if $K$ is a kernel, $L$ must be positive semidefinite for any set of $n$ points $x^{(i)}$. It turns out that it is is also a sufficent condition.
Theorem. (Mercer) Let $K: \mathcal{X} \times \mathcal{X} \to [0,\infty]$ be a kernel function. There exists a mapping $\phi$ associated with $K$ if for any $n$ and any dataset $\{x^{(1)}, x^{(2)}, \ldots, x^{(n)}\}$ of size $n \geq 1$, if and only if the matrix $L$ defined as $L_{ij} = K(x^{(i)}, x^{(j)})$ is symmetric and positive semidefinite.
This characterizes precisely which kernel functions correspond to some $\phi$.
Are kernels a free lunch? Not quite.
Examples of algorithms in which we can use kernels include:
Kernels are very powerful because they can be used throughout machine learning.