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Let $A$ be an $n\times n$ matrix, with $A_{ij}=i+j$. Find the eigenvalues of $A$. A student that I tutored asked me this question, and beyond working out that there are 2 nonzero eigenvalues $a+\sqrt{b}$ and $a-\sqrt{b}$ and $0$ with multiplicity $n-2$, I'm at a bit of a loss.

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    $\begingroup$ What else do you need? $\endgroup$
    – Demosthene
    May 1, 2015 at 19:41
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    $\begingroup$ The eigenvalues seem to be $T(n+1)\pm\sqrt{\dfrac{(n+1)^2(n+2)(2n+3)}{6}}$ and $0$, where $T$ is the triangular numbers sequence. Maybe you can try induction. $\endgroup$
    – Git Gud
    May 1, 2015 at 19:45
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    $\begingroup$ It's just a matrix of the form $uv^T+vu^T$. See this question. $\endgroup$
    – user1551
    May 1, 2015 at 20:18
  • $\begingroup$ In particular, take $u = (1,\dots,1)$ and $v = (1,2,\dots,n)$ $\endgroup$ May 1, 2015 at 20:37

1 Answer 1

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Let $u=(1,1,\ldots,1)^T$ and $v=(1,2,\ldots,n)^T$. Then $A=uv^T+vu^T$. As $u$ is not parallel to $v$ and they are nonzero vectors, $A$ has rank 2 and every eigenvector for a nonzero eigenvalue $\lambda$ must lie in the span of $u$ and $v$. Let $(\lambda,\,pu+qv)$ be such an eigenpair. Then $$ \lambda(pu+qv)=(uv^T+vu^T)(pu+qv)=u[v^T(pu+qv]+v[u^T(pu+qv)]. $$ Since $u$ and $v$ are linearly independent, by comparing coefficients on both sides, we get $$ \lambda\pmatrix{p\\ q}=\underbrace{\pmatrix{v^Tu&v^Tv\\ u^Tu&u^Tv}}_B\pmatrix{p\\ q}. $$ Therefore the nonzero eigenvalues of $A$ are exactly the two eigenvalues of $B$. Since the characteristic polynomial of $B$ is $$ \lambda^2-2(u^Tv)\lambda+[(u^Tv)^2-(u^Tu)(v^Tv)] \equiv (\lambda-u^Tv)^2-(u^Tu)(v^Tv), $$ it follows that $$ \color{red}{\lambda=u^Tv\pm\sqrt{(u^Tu)(v^Tv)}.} $$

In your case, by putting $u=(1,1,\ldots,1)^T$ and $v=(1,2,\ldots,n)^T$, we get $$ \lambda = \frac{n(n+1)}2\pm\sqrt{\frac{n^2(n+1)(2n+1)}6}. $$

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