Eigenvalues of a Complete graph

The eigenvalues should be $n-1$, with multiplicity $1$, and $-1$, with multiplicity $n-1$. The best way to see this in this particular case is through explicitly giving the eigenvectors.

First, the graph $K_{n}$ is $(n-1)$-regular; a $k$-regular graph always has $k$ as an eigenvalue with eigenvector $j$ (the all-ones vector). All other eigenvectors will have to be orthogonal to $j$.

To find the other eigenvectors, consider the adjacency matrix $A$ of $K_{n}$; it is all $1$s, except with $0$ on the diagonal. If we consider $A+I$, we get the all-ones matrix, which has rank $1$ (and so its null space has dimension $n-1$, giving $n-1$ linearly independent eigenvectors for the eigenvalue $-1$). These vectors all take the form $e_{i}-e_{j}$ for $i \neq j$ (where $e_{i}$ represents the vector with a $1$ in position $i$ and a $0$ everywhere else). We can find $n-1$ linearly independent vectors of this type, it is easiest to just consider $e_{1} - e_{j}$ for $0 < j \leq n$.

It's worth adding that the eigenvalues of the Laplacian matrix of a complete graph are $0$ with multiplicity $1$ and $n$ with multiplicity $n-1$.

Recall that the Laplacian matrix for graph $G$ is

$L_G = D - A$

where $D$ is the diagonal degree matrix of the graph. For $K_n$, this has $n-1$ on the diagonal, and $-1$ everywhere else. The constant vector $\mathbf{1}$ is then an eigenvector with eigenvalue $0$. The vectors $v^i$ with $v^i_0 = -1$ and $v^i_i = 1$ for $i \in [2,n]$ give the remaining eigenvectors with eigenvalue $n$.