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Prove that every bilinear form $f:\mathbb R^n \times \mathbb R^n\rightarrow \mathbb R$ has a basis $\{v_1,\ldots,v_n\} \subset \mathbb R^n$ such that $f(v_i,v_j)=-f(v_j,v_i)$ for every $i\neq j$.

I have no idea how to start thinking about this one. Should I actually "build" the required basis or what? Any ideas ?

thanks

MPW
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Kob1
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    I think this is not true, you can consider any simmetric bilinear form as a counterexample. – DiegoMath Jun 20 '14 at 21:05
  • @DiegoMath But that's not a counterexample, take any basis you like for $\mathbb{R}$ and the condition is satisfied because it is vacuous ($f(v_i,v_j) = -f(v_j, v_i)$ for all $i \ne j,\ i,j\in{1}$). – Alex J Best Jun 20 '14 at 21:09
  • Try this counterexample: $f:\mathbb{R^n}\times\mathbb{R^n}\to\mathbb{R}$, defined by $f(x,y)=\langle x,y\rangle$. This is a bilinear form such that $f(x,y)=f(y,x)$, $\forall x,y\in\mathbb{R^n}$. – DiegoMath Jun 20 '14 at 21:14
  • @DiegoMath I was pointing out that in your original counterexample the the condition was fine as there was only 1 dimension and so there is no distinct pair $i,j$ both in ${1}$ for which anything would need to hold. – Alex J Best Jun 20 '14 at 21:18
  • Let us continue this discussion in chat. – Alex J Best Jun 20 '14 at 21:29
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    Where does this question actually come from? – Will Jagy Jun 20 '14 at 22:08
  • Given any bilinear form $f$, consider its symmetric part

    $$f_{sym} : \mathbb{R}^n \times \mathbb{R}^n \ni (u,v) \quad\mapsto\quad f_{sym}(u,v) = \frac12 [ f(u,v) + f(v,u) ] \in \mathbb{R}$$

    Pick an arbitrary basis $\big{;u_1, u_2, \ldots, u_n;\big}$ of $\mathbb{R}^n$ and define the matrix $A = (A_{ij})$:

    $$A_{ij} = f_{sym}( u_i, u_j ),\quad 1 \le i, j \le n$$

    $A$ is a real symmetric matrix and hence diagonalizable through orthogonal matrices.
    i.e. there exists $O = (O_{ij}) \in O(n)$ such that $D = OAO^{-1} = OAO^T$ is a diagonal matrix.

     – achille hui Jun 21 '14 at 02:49
  • Construct a new basis $\big{;v_1, v_2,\ldots,v_n;\big}$ for $\mathbb{R}^n$:

    $$v_i = \sum_{j=1}^n O_{ij} u_j,\quad i = 1,2,\ldots, n$$

    With respect to this basis, we have

    $$f_{sym}( v_i, v_j ) = f_{sym}\left( \sum_{i'=1}^n O_{ii'} u_{i'}, \sum_{j'=1}^n O_{jj'} u_{j'} \right) = (OAO^T){ij} = D{ij} $$ When $i \ne j$, $D_{ij} = 0$ and hence $f_{sym}( v_i, v_j ) = 0 \iff f(v_i,v_j) = - f(v_j,v_i)$.

     – achille hui Jun 21 '14 at 02:50

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