Good day, I'm having a bit of trouble with this one, I'm given the matrix \begin{bmatrix}1&2&...&n\\n+1&n+2&... &2n\\...&...&...&...\\n^2-n+1&n^2-n+2&...&n^2\end{bmatrix} And I'm asked to compute the characteristic polynomial, with the added hint that $W=<(1,1,...,1),(1,2,...,n)>$ is invariant.
I'm a bit confused by the relationship between rows, any help will be appreciated.
you have matrix $$ u v^T + n (v-u) u^T $$ where $$ u = \left( \begin{array}{c} 1 \\ 1 \\ 1 \\ ... \\ 1 \end{array} \right) $$ $$ v = \left( \begin{array}{c} 1 \\ 2 \\ 3 \\ ... \\ n \end{array} \right) $$ where $u \cdot v \neq 0$
We can make a basis $u,v,w_3,w_4, ..., w_n$ where all $w_i \cdot u = 0, \; $ all $w_i \cdot v = 0, \; $ and $i \neq j$ gives $w_i \cdot w_j = 0.$ It follows that all $w_i$ are eigenvectors with eigenvalue $0. \;$
We just need to find out what happens to vectors $a u + b v,$ that is $$ \left( u v^T + n (v-u) u^T \right) (au + bv) = ?? $$ and finding these final eigenvectors and eigenvalues. We know, for real $a,b,$ that the result is some $cu+dv$ with $c,d$ also real. So we expect to get two final real eigenvalues.
Using $$ E = u \cdot u = n, \; \; F = u \cdot v = \frac{n(n+1)}{2}, \; \; G = v \cdot v = \frac{n(n+1)(2n+1)}{6} \; \; , $$ we want the eigenvalues of $$ \left( \begin{array}{cc} F - n E & G - n F \\ n E & nF \end{array} \right) $$ I guess the actual nonzero eigenvalues are a bit of a mess, however the characteristic polynomial comes out $$ x^n - \left( \frac{n(n^2+1)}{2} \right) x^{n-1} - \left( \frac{n^3(n^2-1)}{12} \right) x^{n-2} $$