$n \times n$ matrix, $n-$1 on diagonal, $-1$ off diagonal, find eigenvalues

Question

just wanted to pose a follow-up to the previous question I asked at Symmetric matrix, $a>0$ on diagonal and $b<0$ off diagonal, positive definite?:

Suppose we have an $n \times n$ matrix whose diagonal entries are $n-1$ and whose non-diagonal entries are $-1$. Show that the eigenvalues are 0 (multiplicity $1$) and $n$ (multiplicity $n-1$).

I got the following comment:

The vector $v$ with constant coefficients equal to $1$ is eigen. It corresponds to the eigenvalue $0$. As your matrix is symmetric, it is diagonalizable on an orthonormal basis. Set $V=v^\perp$. Check that $OMO^{-1}=M$ for every $O\in O(n)$ that let $v$ (thus $V$) stable. This implies $M$ as a single eigenvalue on $V$, which thus has multiplicity $n-1$. Another way to do it is to verify that if $w\in V$ is eigen, then every permutation of its components is also eigen. A third way is to write your matrix as $cI-dE$ where $E$ is full $1$, then use the characteristic polynomial.

Sorry, but I don't understand this answer. I got that $(1,1,...1)$ is an eigenvector with eigenvalue $0$. The orthogonal complement to this will be $\{(x_1,...,x_n)\in\mathbb{R}^n:x_1+...+x_n=0\}$. But from here it gets confusing.

• What is $M$? How would $OMO^{-1}=M$ for every $O\in O(n)$ (I assume that is the group of orthogonal matrices) imply that there is only one other eigenvalue?
• I assume he meant any permutation of the components of an eigenvector is also an eigenvector $\mathit{\text{with the same eigenvalue}}$. That's easy to show. But it still leaves the question of finding an eigenvector such that the subspace generated by its set of permutations of components has dimension $n-1$. How do I do that?
• In this case, that means $(n-1)I-E$. But to find the characteristic polynomial from here... Brute force?

Thanks

Answer 1

What the question promises you is that there is an $n-1$ dimensional space with all the vectors in it eigenvectors. That is clearly the space perpendicular to $(1,1,1,\ldots 1)$. A handy vector in that space is $(1-n,1,1,1,\ldots 1)$. Can you prove that is an eigenvector with eigenvalue $n$? Now the symmetry of the problem says you can rotate the entries of the vector to put the $1-n$ in any location and it will still be an eigenvector. You should be able to prove these vectors span the space of interest.

Answer 2

not sure what you are after, but call your symmetric matrix $M$ for $n=5,$ here is a matrix $P$ with the columns all being eigenvectors and all perpendicular to each other. You can make my matrix orthogonal by dividing each column by its length, which is a square root,.,, Note that $P^T MP$ is diagonal, although the diagonal entries may not be the exact eigenvalues unless you normalize the columns of $P$

$$ \left( \begin{array}{rrrrr} 1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 \\ 1 & 0 & 2 & -1 & -1 \\ 1 & 0 & 0 & 3 & -1 \\ 1 & 0 & 0 & 0 & 4 \\ \end{array} \right). $$