I have to prove that if $V$ is a real vector space ($\dim V=n$) with inner product $(.,.)$ then if we define $$ (v_{1}\wedge v_{2}\wedge\cdots\wedge v_{k},w_{1}\wedge w_{2}\wedge\cdots\wedge w_{k}) =\det((a_{j,r})_{j,r=1,\dots,k}), $$ where $a_{j,r}=(v_{j},v_{r})$, and we extend it by linearity it defines an inner product on $\Lambda^{k}V$. Same question in the hermitian case.

I have proved that $(v_{1}\wedge v_{2}\wedge...\wedge v_{k},v_{1}\wedge v_{2}\wedge...\wedge v_{k}) \ge0 $ and it is zero iff $v_{1}\wedge v_{2}\wedge...\wedge v_{k} = 0$. but what about a generic element of $\Lambda^{k}V$ that is linear combination of "simple" elements? thank you!

  • $\begingroup$ Use an orthonormal basis. $\endgroup$ – darij grinberg Nov 18 '14 at 20:28
  • $\begingroup$ can you be more explicit please?I have to consider an orthonormal basis $\{e_{i}\} $ and then prove that $(e_{i_{1}}\wedge e_{i_{2}}\wedge...\wedge e_{i_{k}}, i_{1}<i_{2}..<i_{k} ) is an orthonormal basis of the exterior algebra? $\endgroup$ – joker Nov 18 '14 at 20:38
  • $\begingroup$ Yes! (Or "orthogonal" both times -- this is a bit better.) $\endgroup$ – darij grinberg Nov 18 '14 at 21:22
  • $\begingroup$ so $$ (e_{i_{1}}\wedge e_{i_{2}}\wedge...\wedge e_{i_{k}},e_{i_{1}}\wedge e_{i_{2}}\wedge...\wedge e_{i_{k}}) =det(identity matrix KxK) $$ $\endgroup$ – joker Nov 18 '14 at 22:01
  • $\begingroup$ but why $$(e_{i_{1}}\wedge e_{i_{2}}\wedge...\wedge e_{i_{k}},e_{j_{1}}\wedge e_{j_{2}}\wedge...\wedge e_{j_{k}})=0 $$? $\endgroup$ – joker Nov 18 '14 at 22:02

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