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So on a complete graph of $4$ vertices we have

$$ A= \begin{bmatrix} 0 & 1 & 1 & 1 \\ 1& 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 &0 \end{bmatrix} $$

And we want to find all eigenvalues so have to compute det($A- \lambda I$) so we would have $$ A= \begin{bmatrix} -\lambda & 1 & 1 & 1 \\ 1& -\lambda & 1 & 1 \\ 1 & 1 & -\lambda & 1 \\ 1 & 1 & 1 & -\lambda \end{bmatrix} $$

I used wolfram alpha, And I got

$$\lambda = 3, v_1 = (1,1,1,1)$$ $$\lambda = -1, v_2 = (-1,0,0,1), v_3 = (-1,0,1,0), v_4 = (-1,1,0,0)$$

And I also want to generalize this question to a complete graph on $n$ Vertices.

Would I just take the positive value of $\lambda =3$ and so in general, $\lambda = n-1$, Does this make sense ? But why would I disregard all other values ?

alkabary
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2 Answers2

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The answer is simple:

If $G$ is $n\times n$ complete graph, then $$A(G) = J_n - I_n$$ Where $J_n$ is the $n\times n$ square matrix of all ones. Since $J_n$ and $I_n$ commutes. Then: $$spect(A(G)) = spect(J_n) - spect(I_n) \;(to\; certain\; order)$$ It's easy to see that $spect(J_n)= (n,0,\ldots,0)$ Then $$spect(A(G)) = (n-1,-1,\ldots,-1)$$

M.Badaoui
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$A+I$ is a rank-$1$ matrix with eigenvalues $\{4,0,0,0\}$, hence the eigenvalues of $A$ are $\{3,-1,-1,-1\}$. In general, the eigenvalues of the $n\times n$ matrix with $a_{ij}=1-\delta(i,j)$ are $n-1$ with multiplicity one and $-1$ with multiplicity $n-1$.

Jack D'Aurizio
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  • So for a complete graph on $5$ vertices, I would have ${4,-1,-1,-1,-1 }$ and so on ? – alkabary Oct 29 '15 at 19:32
  • @alkabary: exactly. – Jack D'Aurizio Oct 29 '15 at 19:33
  • @alkabary, I found a nice square matrix with columns that are a basis of eigenvectors for the matrix with all entries $1.$ It therefore works for any multiple of that plus any multiple of the identity. Anyway, columns orthogonal to each other, just not adjusted to common length: http://math.stackexchange.com/questions/1414690/matrix-with-all-1s-diagonalizable-or-not/1414925#1414925 For you, take the upper left 4 by 4 corner square. Check! – Will Jagy Oct 29 '15 at 20:13