= Solution
A <matrix> is a <diagonalizable matrix> if there is an invertible matrix $P$ for which $P^{-1}MP$ is diagonal. Its columns then form a <basis> of <eigenvectors>, and the diagonal entries are their corresponding <eigenvalues>.
Here the given eigenvectors form an <orthonormal basis>. Put them into the columns of $Q$:
$$
Q=(\mathbf e_1\ \cdots\ \mathbf e_n),\qquad Q^TQ=I,\qquad Q^{-1}=Q^T.
$$
Define $\Lambda=Q^TMQ$. Its entries are
$$
\Lambda_{ab}=\mathbf e_a^TM\mathbf e_b
=\lambda_b\,\mathbf e_a\cdot\mathbf e_b
=\lambda_b\delta_{ab}.
$$
This proves the claimed <diagonalization of a matrix>, rather than merely guessing it:
$$
\boxed{\Lambda=\operatorname{diag}(\lambda_1,\ldots,\lambda_n),\qquad
M=Q\Lambda Q^T.}
$$
The <trace> follows directly from this factorization:
$$
\operatorname{tr}M
=\sum_i\sum_a\lambda_aQ_{ia}^2
=\sum_a\lambda_a\sum_iQ_{ia}^2
=\sum_a\lambda_a,
$$
because each column of $Q$ has unit length. \b[Trace is the sum of the eigenvalues, with multiplicity.]
For the particular matrix, $(M^2)_{ij}=\sum_kM_{ik}M_{kj}$. If $i=j$, every $k\ne i$ contributes one and $k=i$ contributes zero. If $i\ne j$, precisely $k=i$ and $k=j$ contribute zero; the other $n-2$ terms contribute one. Thus
$$
(M^2)_{ij}=
\begin{cases}
n-1,&i=j,\\
n-2,&i\ne j,
\end{cases}
\qquad
M^2=(n-1)I+(n-2)M.
$$
Applying this identity to a nonzero <eigenvector> gives
$$
\lambda^2=(n-1)+(n-2)\lambda,
\qquad
(\lambda-(n-1))(\lambda+1)=0.
$$
Every <eigenvalue> is therefore $n-1$ or $-1$.
If $r$ is the multiplicity of $n-1$, the <trace> formula for a <diagonalizable matrix> and the zero diagonal give
$$
0=r(n-1)+(n-r)(-1)=n(r-1),
$$
so $r=1$. Up to ordering, the required diagonal form is
$$
\boxed{\Lambda=\operatorname{diag}(n-1,-1,\ldots,-1).}
$$
For $n=1$ this means just $(0)$. For $n\ge2$, the eigendirections are also transparent: the all-ones vector has eigenvalue $n-1$, and the $(n-1)$-dimensional subspace whose coordinate sum is zero has eigenvalue $-1$, since $(M\mathbf x)_i=\sum_jx_j-x_i$. \b[The spectrum consists of one collective mode and $n-1$ zero-sum modes.]
Back to article page
