# When the Lie algebra of matrices with zero last rows is Frobenius?

Let $\mathcal{A}_{n,k}$ be the Lie algebra of $n \times n$ matrices over $\mathbb{C}$ for which the last $k$ rows are equal to zero. Suppose that $k$ does not divide $n$. How to prove that $\mathcal{A}_{n,k}$ is not a Frobenius Lie algebra, i.e., that for every form $l \in \mathcal{A}_{n,k}^*$ the bilinear form $\langle a, b\rangle = l([a, b])$ is degenerate?

(For the case when $k$ divides $n$ I know the proof that it is Frobenius.)

• And how does the proof that you know go? – darij grinberg Jun 3 '15 at 12:35
• One can construct a certain non-degenerate form $l$ when $k$ divides $n$. See arxiv.org/pdf/1310.1193v1.pdf for example. – Alexey Jun 3 '15 at 13:03
• Hmm. Your $\mathcal{A}_{N, M}$ there is a different Lie algebra. Are they conjugate? – darij grinberg Jun 3 '15 at 13:20
• Yes, they are isomorphic. – Alexey Jun 3 '15 at 13:24