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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!

user26857
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joker
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  • Use an orthonormal basis. – darij grinberg Nov 18 '14 at 20:28
  • 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? – joker Nov 18 '14 at 20:38
  • Yes! (Or "orthogonal" both times -- this is a bit better.) – darij grinberg Nov 18 '14 at 21:22
  • so $$ (e_{i_{1}}\wedge e_{i_{2}}\wedge...\wedge e_{i_{k}},e_{i_{1}}\wedge e_{i_{2}}\wedge...\wedge e_{i_{k}}) =det(identity matrix KxK) $$ – joker Nov 18 '14 at 22:01
  • but why $$(e_{i_{1}}\wedge e_{i_{2}}\wedge...\wedge e_{i_{k}},e_{j_{1}}\wedge e_{j_{2}}\wedge...\wedge e_{j_{k}})=0 $$? – joker Nov 18 '14 at 22:02
  • Just look at the matrix. I'm claiming that it always has a row full of zeroes unless $\left(i_1,i_2,...,i_k\right) = \left(j_1,j_2,...,j_k\right)$. – darij grinberg Nov 18 '14 at 22:04
  • I am working on (almost) the same problem and am farther behind than you in understanding it. (It's one of my first problems involving exterior algebra). Would you mind sharing your proof that the thing-we-hope-is-an-inner-product is positive-definite? I'm hoping that will help me understand how the definitions fit together. Thank you. – stwwilkinson Oct 15 '16 at 04:37

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