# Inner product exterior algebra

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!

• 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