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We have a bilinear form,

$$\langle A, A' \rangle = \det(A+A') - \det A - \det A'$$

on the real vector space of Hermitian $2\times 2$ matrices.

If I have a basis of the space, I would calculate the norms of the basis vectors, but here we don't. What should I do?

Jean Marie
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user398843
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  • Why do you say that it is bilinear? – ancient mathematician May 08 '19 at 06:46
  • @ancientmathematician One can compute it by using the identity $\det(A+B) = \det(A) +\det(B) +tr(A^*B)$ – user398843 May 08 '19 at 06:49
  • Of course, sorry, half-asleep. The space is surely four dimensional with basis , $\begin{pmatrix}1 & 0 \ 0 & 0 \end{pmatrix}$,$\begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}$,$\begin{pmatrix}0 & 1 \ 1 & 0 \end{pmatrix}$,$\begin{pmatrix} 0 & i \ -i & 0 \end{pmatrix}$ so why not calculate? – ancient mathematician May 08 '19 at 07:15
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    @ancientmathematician the result beeing $(3+, 1-)$, i.e. a minkovskian space (space-time metrics) – Jean Marie May 08 '19 at 07:29
  • Related : see some interesting answers and comments in https://math.stackexchange.com/q/1642024 – Jean Marie May 08 '19 at 08:10
  • I have taken the liberty to modify your title in order to attract more people. I have changed as well the tag ("abstract algebra" is for structures like groups, rings, fields) – Jean Marie May 08 '19 at 08:42
  • Relate as well: https://math.stackexchange.com/q/405276 – Jean Marie May 08 '19 at 08:49

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