I'm studying the proof of the following statement:
$Spec(K_n) = (n-1)^1(-1)^{n-1}$
At some point I have:
By the Spectral Theorem, when looking for eigenvectors $v$ we can assume they are orthogonal to $j$.
$j$ being the all-1 vector.
I don't understand which part of the Spectral Theorem implies this?
The spectral thoerem tells us, that a symmetric matrix (e. g. the adjacency matrix $A$ of the $K_n$) has a orthogonal basis of eigenvectors. As $j$ is an eigenvector for the eigenvalue $n-1$ (since each row of $A$ has row sum $n-1$), when looking for eigenvectors for $-1$ we must concentrate on $\{j\}^\perp$.