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Let $F$ be a field, for $H \in M_{k \times k}(F)$, let $H^*$ be the adjugate matrix of $H$. (Where the adjugate $H^*$ is the transpose of the matrix of cofactors of $H$.)

I am trying to prove the following two results:

1) If $H, G \in M_{n \times n}(F)$, then ${(HG)}^* = G^* H^*$.

2) If $H \in M_{n \times n}(F)$ and $G \in M_{m \times m}(F)$, then \begin{equation*} \begin{pmatrix} H & 0 \\ 0 & G \end{pmatrix}^* = \begin{pmatrix} \det(G) H^* & 0 \\ 0 & \det(H) G^* \end{pmatrix}. \end{equation*}

My current idea, is that both of these results hold for invertible matrices (using that $H^* = \det(H) H^{-1}$, when $H \in GL_n(F)$). Then, using that the set of invertible matrices is Zariski dense $M_{n \times n}(F)$, deduce the result for all matrices.

This argument is very similar to the topological proof of the Cayley-Hamilton theorem. To this end it would be useful to have a reference for a proof of the Cayley-Hamilton theorem using algebraic geometry. Alternatively, does anyone know of a reference which proves the two results.

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closed as off-topic by Andreas Blass, Paul Siegel, Ben McKay, Peter Heinig, Olivier Feb 26 '18 at 17:49

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  • $\begingroup$ $H^*$ is uniquely determined by the identity $H H^* = H^* H = \det(H) I$. I think both of your formulas follow from that (together with the usual properties of determinants). $\endgroup$ – Paul Siegel Feb 26 '18 at 15:57
  • $\begingroup$ @PaulSiegel I agree. It's a reference for the proof that both my formulas follow from this property that I'm looking for. $\endgroup$ – David Watson Feb 26 '18 at 16:30
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    $\begingroup$ @PaulSiegel, of course I guess your comment means "the map $H \mapsto H^*$ is uniquely determined by the fact that it is Zariski continuous and …". $\endgroup$ – LSpice Feb 26 '18 at 16:49