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.

  • 1
    $\begingroup$ What else do you need? $\endgroup$
    – Demosthene
    May 1, 2015 at 19:41
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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


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 $$ \lambda = \frac{n(n+1)}2\pm\sqrt{\frac{n^2(n+1)(2n+1)}6}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.