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Let $tr(A,B)$ be a bilinear form on $\mathcal{M}(n,\mathbb{K})$. It is obviously symmetric. But, how can I find the signature and the rank?

I want to construct the matrix associated to the above bilinear form. I think it has 1 in positions $(1,1)$, $(2,n+1)$, $(3,2n+1),\dots,(i,in+1)$ and their symmetric entries. Hence, the rank is $n^2$. But what about the signature? It is not easy to find the eigenvalues he

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    By Sylvester's theorem, it is enough to do the calculations on **any** basis of $\;\mathcal M(n,\Bbb K)\;$ ... – 2017-02-18
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    Which basis I have to choose? – 2017-02-18
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    How about the simplest you can possibly think of? – 2017-02-18
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    The canonic one, but the matrix is represented in the way I described above. – 2017-02-18

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Hint. Note that $\operatorname{tr}(E_{ij}B)$ is nonzero only if $B$ is a multiple of $E_{ij}^T$. Reorder the canonical basis of $M_n(\mathbb K)$ so that

  • the strictly upper/lower triangular $E_{ij}$s precede all diagonal $E_{ii}$s,
  • each strictly upper triangular $E_{ij}$ is immediately followed by its transpose, and
  • the subset of all strictly upper triangular $E_{ij}$s are ordered lexicographically.

For instance, when $n=3$, reorder the basis as $$ E_{12},E_{12}^T,E_{13},E_{13}^T,E_{23},E_{23}^T,E_{11},E_{22},E_{33}. $$ What is the matrix representation of the symmetric bilinear form $(A,B)\mapsto\operatorname{tr}(AB)$ with respect to this basis?

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    (+1) Nice solution. So, to state it clearly, the rank is $n^2$ and the signature is $$ (+1)^{\frac{n^2+n}{2}},\quad (-1)^{\frac{n^2-n}{2}}.$$ – 2017-07-06