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Let $p, q > 0$ be integers and let $K_{p,q}$ be a complete bipartite graph. Let $A(K)$ denote the adjacency matrix of $K_{p,q}$ according to a convenient labeling of vertices , which is the $2 \times 2$ block matrix

$$\begin{bmatrix} 0_{p \times p} & 1_{p \times q}\\ 1_{q \times p} & 0_{q \times q} \end{bmatrix}$$

I know that the spectral radius of $A(K)$ is $\sqrt{pq}$ but couldn't prove it. I tried to find the determinant of the characteristic polynomial in order to find the spectrum but also couldn't. Any help would be appreciated.

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1 Answer 1

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The spectral radius of $A(K)$ is the square root of the spectral radius of $A(K)^2$. Note that $$ A(K)^2 = \begin{pmatrix} q\mathbf{1}_{p \times p} & 0\\ 0 & p\mathbf{1}_{q \times q} \end{pmatrix}. $$ Hence $$ \det(xI - A(K)^2) = \det(xI - q\mathbf{1}_{p \times p})\det(xI - p\mathbf{1}_{q \times q}) = (x - pq)^2x^{p + q - 2}. $$

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  • $\begingroup$ Define $X$ please. $\endgroup$
    – emelie
    Commented Apr 15, 2018 at 12:20
  • $\begingroup$ @sal.f Sorry, I meant $K$. Edited. $\endgroup$ Commented Apr 15, 2018 at 12:22
  • $\begingroup$ No worries, thank you. $\endgroup$
    – emelie
    Commented Apr 15, 2018 at 12:27
  • $\begingroup$ Sorry, but how did you find $\det(xI - q\mathbf{1}_{p \times p})$ ? $\endgroup$
    – emelie
    Commented Apr 15, 2018 at 12:57
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    $\begingroup$ $pq$ is clearly is an eigenvalue of $q\mathbf{1}_{p \times p}$ (take $\mathbf{1}_{p \times 1}$ as an eigenvector) and this matrix has rank 1, so $0$ is an eigenvalue with multiplicity $p - 1$, which implies that $pq$ has multiplicity $1$ and there can be no other eigenvalues. $\endgroup$ Commented Apr 15, 2018 at 13:13

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