I saw an LDA (linear discriminant analysis) plot with decision boundaries from The Elements of Statistical Learning:
I understand that data are projected onto a lowerdimensional subspace. However, I would like to know how we get the decision boundaries in the original dimension such that I can project the decision boundaries onto a lowerdimensional subspace (likes the black lines in the image above).
Is there a formula that I can use to compute the decision boundaries in the original (higher) dimension? If yes, then what inputs does this formula need?
Answer
This particular figure in Hastie et al. was produced without computing equations of class boundaries. Instead, algorithm outlined by @ttnphns in the comments was used, see footnote 2 in section 4.3, page 110:
For this figure and many similar figures in the book we compute the decision boundaries by an exhaustive contouring method. We compute the decision rule on a fine lattice of points, and then use contouring algorithms to compute the boundaries.
However, I will proceed with describing how to obtain equations of LDA class boundaries.
Let us start with a simple 2D example. Here is the data from the Iris dataset; I discard petal measurements and only consider sepal length and sepal width. Three classes are marked with red, green and blue colours:
Let us denote class means (centroids) as $\boldsymbol\mu_1, \boldsymbol\mu_2, \boldsymbol\mu_3$. LDA assumes that all classes have the same withinclass covariance; given the data, this shared covariance matrix is estimated (up to the scaling) as $\mathbf{W} = \sum_i (\mathbf{x}_i\boldsymbol \mu_k)(\mathbf{x}_i\boldsymbol \mu_k)^\top$, where the sum is over all data points and centroid of the respective class is subtracted from each point.
For each pair of classes (e.g. class $1$ and $2$) there is a class boundary between them. It is obvious that the boundary has to pass through the middlepoint between the two class centroids $(\boldsymbol \mu_{1} + \boldsymbol \mu_{2})/2$. One of the central LDA results is that this boundary is a straight line orthogonal to $\mathbf{W}^{1} \boldsymbol (\boldsymbol \mu_{1} – \boldsymbol \mu_{2})$. There are several ways to obtain this result, and even though it was not part of the question, I will briefly hint at three of them in the Appendix below.
Note that what is written above is already a precise specification of the boundary. If one wants to have a line equation in the standard form $y=ax+b$, then coefficients $a$ and $b$ can be computed and will be given by some messy formulas. I can hardly imagine a situation when this would be needed.
Let us now apply this formula to the Iris example. For each pair of classes I find a middle point and plot a line perpendicular to $\mathbf{W}^{1}
\boldsymbol (\boldsymbol \mu_{i} – \boldsymbol \mu_{j})$:
Three lines intersect in one point, as should have been expected. Decision boundaries are given by rays starting from the intersection point:
Note that if the number of classes is $K\gg 2$, then there will be $K(K1)/2$ pairs of classes and so a lot of lines, all intersecting in a tangled mess. To draw a nice picture like the one from the Hastie et al., one needs to keep only the necessary segments, and it is a separate algorithmic problem in itself (not related to LDA in any way, because one does not need it to do the classification; to classify a point, either check the Mahalanobis distance to each class and choose the one with the lowest distance, or use a series or pairwise LDAs).
In $D>2$ dimensions the formula stays exactly the same: boundary is orthogonal to $\mathbf{W}^{1} \boldsymbol (\boldsymbol \mu_{1} – \boldsymbol \mu_{2})$ and passes through $(\boldsymbol \mu_{1} + \boldsymbol \mu_{2})/2$. However, in higher dimensions this is not a line anymore, but a hyperplane of $D1$ dimensions. For illustration purposes, one can simply project the dataset to the first two discriminant axes, and thus reduce the problem to the 2D case (that I believe is what Hastie et al. did to produce that figure).
Appendix
How to see that the boundary is a straight line orthogonal to $\mathbf{W}^{1} (\boldsymbol \mu_{1} – \boldsymbol \mu_{2})$? Here are several possible ways to obtain this result:

The fancy way: $\mathbf{W}^{1}$ induces Mahalanobis metric on the plane; the boundary has to be orthogonal to $\boldsymbol \mu_{1} – \boldsymbol \mu_{2}$ in this metric, QED.

The standard Gaussian way: if both classes are described by Gaussian distributions, then the loglikelihood that a point $\mathbf x$ belongs to class $k$ is proportional to $(\mathbf x – \boldsymbol \mu_k)^\top \mathbf W^{1}(\mathbf x – \boldsymbol \mu_k)$. On the boundary the likelihoods of belonging to classes $1$ and $2$ are equal; write it down, simplify, and you will immediately get to $\mathbf x^\top \mathbf W^{1} (\boldsymbol \mu_{1} – \boldsymbol \mu_{2}) = \mathrm{const}$, QED.

The laboursome but intuitive way. Imagine that $\mathbf{W}$ is an identity matrix, i.e. all classes are spherical. Then the solution is obvious: boundary is simply orthogonal to $\boldsymbol \mu_1 – \boldsymbol \mu_2$. If classes are not spherical, then one can make them such by sphering. If the eigendecomposition of $\mathbf{W}$ is $\mathbf{W} = \mathbf U \mathbf D \mathbf U^\top$, then matrix $\mathbf S = \mathbf D^{1/2} \mathbf U^\top$ will do the trick (see e.g. here). So after applying $\mathbf S$, the boundary is orthogonal to $\mathbf S (\boldsymbol \mu_{1} – \boldsymbol \mu_{2})$. If we take this boundary, transform it back with $\mathbf S^{1}$ and ask what is it now orthogonal to, the answer (left as an exercise) is: to $\mathbf S^\top \mathbf S \boldsymbol (\boldsymbol \mu_{1} – \boldsymbol \mu_{2})$. Plugging in the expression for $\mathbf S$, we get QED.
Attribution
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