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. $$ So the eigenvalues of $C_1$ are the eigenvalues of the $\hat A_n$, which in this $2 \times 2$ case are given by the formula $\frac12(t \pm \sqrt{t^2-4\Delta})$ where $t$ is the trace and $\Delta$ … Compute the eigenvalues of this circulant matrix $C$ as usual, and then extract their square roots to recover the eigenvalues of $C_1$. …

For the "complementary" problem, $g(n)<n$ for all positive $n$, and the upper bound $g(n) \leq n-1$ is sharp at least for $n=1,2,4,8$. If $\dim W \geq n$ then any nonzero vector $v \in {\bf R}^n$ is …

Hope this isn't a contest/homework problem… Let $v=(1,1,1,\ldots,1)$. Then $(v,Av) = \sum_{i,j} a_{ij} \geq n^2-n = (n-1) (v,v)$. Hence the maximal eigenvalue $\lambda_1$ is at least $n-1$ (Rayleigh …

Counterexample: let $k=7$, and let $B$ be the circulant matrix with $B_{ij}=1$ iff $i-j \in \{1,2,4\} \bmod 7$. Then $X_B$ is $I + \frac12 J$, with characteristic polynomial $(x-1)^6 (x-\frac92)$. O …

$a_i = 1$ then that eigenvalue is $1 + 2 \cos\frac\pi{n+1}$ if I did this right; since you allow only $a_i < 1$, this bound $1 + 2 \cos\frac\pi{n+1}$ is not attained, but it is still the supremum of eigenvalues …

(1) No. Counterexample: the symmetric $3 \times 3$ matrix $$ M(a,b) = \left[ \begin{array}{ccc} 0 & a & b \cr a & 0 & b \cr b & b & 0 \end{array} \right] $$ with $0 < b < a$ has $\lambda_3 = -a$ wit …

Can such a graph be integral, i.e. have only integer eigenvalues? Yes. … Two examples are $(a,b,c,d) = (441, 744, 1225, 5635)$, with eigenvalues $-945$, $-525$, $-3038$, $4058$, and $(a,b,c,d) = (1575, 1900, 4500, 33516)$, with eigenvalues $-1710$, $-2940$, $-14250$, $18900 …