Spielman has some notes where he explains that the eigenvectors of the Laplacian of the ring graph are
$$ x_k(u) = \sin(\frac{2\pi ku}{n})\\y_k(u) = \cos(\frac{2\pi ku}{n}) $$ for $1 \leq k \leq n/2$ where $n$ is the number of vertices in the ring graph. He also notes that $x_k$ is ignored for even $n$ since $x_k$ becomes the 0 vector. However, I thought that the Laplacian of any graph $L$ is guaranteed to have $n$ eigenvectors since $L$ is real-valued and symmetric. What happened to the last eigenvector for the ring graph of even $n$ vertices?
Well, let's try $n=4$. The graph has Laplacian matrix $$ \pmatrix{2 & -1 & 0 & -1\cr -1 & 2 & -1 & 0\cr 0 & -1 & 2 & -1\cr -1 & 0 & -1 & 2\cr} $$ It has eigenvalue $0$ for eigenvector $(1,1,1,1)^T$ (this is Spielman's $y_0$), $2$ for $(1,0,-1,0)$ ($x_1$) and $(0, -1,0,1)$ ($y_1$) and $4$ for eigenvector $(-1,1,-1,1)^T$ ($y_2$). Both $x_0$ and $x_2$ are $(0,0,0,0)^T$. It looks like Spielman neglected to mention $y_0$ in Lemma 2.4.4.