How to determine the eigenvalues and the multiplicity of the eigenvalues of this symmetric matrix?

Question

Suppose we are given a family of $k$ vectors $(\vec{v}_1,\dots,\vec{v}_k)$ in $\mathbb{R}^k$ such that $\vec{v}_i\cdot\vec{v}_i=1$ for all $1\leq i\leq k$ and $||\vec{v}_i-\vec{v}_j||=1$ for all $i\neq j$. How do we compute the eigen values and their multiplicity of the Gram matrix

$$GM(\vec{v}_1,\dots,\vec{v}_k)= \begin{pmatrix} \vec{v}_1\cdot\vec{v}_1 & \vec{v}_1\cdot\vec{v}_2 &\dots &\vec{v}_1\cdot\vec{v}_k\\ \vec{v}_2\cdot\vec{v}_1 & \vec{v}_2\cdot\vec{v}_2 & \dots &\vec{v}_2\cdot\vec{v}_k\\ \vdots & \vdots &\ddots &\vdots\\ \vec{v}_k\cdot\vec{v}_1 & \vec{v}_k\cdot\vec{v}_2&\dots&\vec{v}_k\cdot\vec{v}_k \end{pmatrix} $$

I understand that, by hypothesis, we have $\vec{v}_i\cdot\vec{v}_i$ for all $1\leq i\leq k$ which in turn implies $$\begin{align*} ||\vec{v}_i-\vec{v}_j||^2 =\vec{v}_i\cdot\vec{v}_i - 2(\vec{v}_i\cdot\vec{v}_j)+\vec{v}_j\cdot\vec{v}_j =2-2(\vec{v}_i\cdot\vec{v}_j) \end{align*} $$

so that $$\vec{v}_i\cdot\vec{v}_j=\frac{1}{2}$$ for all $i\neq j$.

This means $$GM(\vec{v}_1,\dots,\vec{v}_k)= \begin{pmatrix} 1 & \frac{1}{2} &\dots & \frac{1}{2}\\ \frac{1}{2} & 1 & \dots &\frac{1}{2}\\ \vdots & \vdots &\ddots &\vdots\\ \frac{1}{2} & \frac{1}{2}&\dots&1 \end{pmatrix} $$

but I am unsure how to continue from here. More generally, is there a method to computing the eigenvales/characteristic polynomials of large symmetric matrices?

Any help would be appreciated. Thank you. (For context, see this paper on page four).

Use https://math.stackexchange.com/questions/899154/characteristic-polynomial-of-a-square-matrix-of-size-n-with-all-entries-equal : if the all 1s matrix is $J$ then your matrix is $\frac{1}{2}(J+I)$ — ancient mathematician, May 06 '21 at 09:19
@ancientmathematician my linear algebra skills aren't too polished yet. do you mind elaborating a bit? — C Squared, May 06 '21 at 09:25
The ev of $J$ are $n$ once and $0$ $n-1$ times. So the ev of $I+J$ are $n+1$ once and $1$ $n-1$ times. So the ev of your matrix are half these. This is standard in adjacency theory of graphs. — ancient mathematician, May 06 '21 at 09:27
@ancientmathematician ok, thank you. i didn't realize that ev's behaved that way with matrix addition — C Squared, May 06 '21 at 09:29
Caution: adding the identity matrix is safe, but it's not true for $A+B$ in general. — ancient mathematician, May 06 '21 at 09:32

score 1 · Accepted Answer · answered May 06 '21 at 10:58

Whenever you have an $n\times n$ matrix of the form $$ A = \begin{pmatrix} a & b &\cdots & b\\ b & a & \cdots &b\\ \vdots & \vdots &\ddots &\vdots\\ b & b&\cdots&a \end{pmatrix} $$ for some non-zero $b$, you can use two observations to determine the eigenvalues and their multiplicities:

The matrix $A-(a-b)I$ has all entries equal to $b$, hence is of rank $1$. This means that $a-b$ is an eigenvalue of geometric multiplicity $n-1$.
All row sums are equal to $a+(n-1)b$, so that the vector $(1,1,\dots,1)^{\textrm t}$ is an eigenvector for the eigenvalue $a+(n-1)b$. The geometric multiplicity has to be $1$ since the other eigenspace is already of dimension $n-1$ and distinct eigenspaces always form direct sums.

How to determine the eigenvalues and the multiplicity of the eigenvalues of this symmetric matrix?

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