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I am having troubles proving that for a symmetric non degenerate bilinear form $\varphi$ on a finite dimensional vector space $V$, with $\dim V>1,$ there is a basis of $V$ $e_1,...,e_n$ such that the matrix $B$ associated to $\varphi$ with respect to $e_1,...,e_n$ is the anti-diagonal matrix. So far I've tried this: Assume that

(a) there exist a vector $x \in V$ different from zero such that $\varphi(x,x)=0$

(b) there exist an $y \in V$ such that $\varphi(x,y)=1$ and $\varphi(y,y)=0$

I try to show the fact by induction on the dimension $n$:

base of induction: $\dim V=2$ choose a basis $e_1,e_2$ for $V$ for assumptions (a) and (b) $\varphi(e_1,e_1)=\varphi(e_2,e_2)=0$ and $\varphi(e_1,e_2)=\varphi(e_2,e_1)=1$.

Inductive Hypothesis: suppose holds for $n-1=\dim V $.

Consider $V$ such that $\dim V=n$ choose a basis $e_1,...,e_n$ for $V$ then for all $x,y \in V$ we have:

$\varphi(x,y)=\varphi(\sum_{i=1}^n x_ie_i,\sum_{j=1}^n y_je_j)=\sum_{i,j=1}^n x_iy_j \varphi(e_i,e_j)= \sum_{i=1}^n x_iy_n\varphi(e_i,e_n) + \sum_{j=1}^n x_n y_j \varphi(e_n,e_j)$ + $\sum_{i,j=1}^{n-1} x_iy_j \varphi(e_i,e_j)$.

By the inductive hypothesis the latter term has a diagonal form, what can I say for the remaining two terms? How to prove that the remaining terms are all zeroes besides the first for the first term? (for the second term should already follow by symmetry)

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  • $\begingroup$ It is false. There is no basis of $\mathbb R^2$ such that $\begin{bmatrix}0&a\\a&0\end{bmatrix}$ represents the Euclidean dot product. $\endgroup$
    – mr_e_man
    May 16, 2023 at 21:14
  • $\begingroup$ As noted by @mr_e_man this result fails for any field. You must add some hypothesis on the field. For example, for algebraically closed fields this result is true. $\endgroup$
    – Daniel
    May 17, 2023 at 0:50

1 Answer 1

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We must assume that the underlying field $k$ is algebraically closed for the existence of your $x$ and $y$ as you defined above.

Define $f_1:V\rightarrow k$, $f_2:V \rightarrow k$ by $f_1(z)=\varphi(x,z)$ and $f_2(z)=\varphi(y,z)$. Define $V'=\ker(f_1)\cap\ker(f_2)$.

Define $F(z)=(f_1(z),f_2(z))$. Notice that $F$ has rank 2, since $F(x)=(0,1), F(y)=(1,0)$, and $\ker(F)=V'$. Thus, $\dim(V')=\dim(V)-2$.

Use induction on the dimension of the space in order to find a basis for $V'$ such that $\varphi$ is anti-diagonal w.r.t. this basis. Let $\{e_2,\ldots,e_{\dim(V)-1}\}$ be this basis. Let $e_1=x$ and $e_{\dim(V)}=y$.

Finally, notice that $\varphi$ w.r.t. $\{e_1,\ldots,e_{\dim(V)}\}$ is anti-diagonal.

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  • $\begingroup$ Thank you, who are f_1 and f_2 ? Do you mean by them the induced isomorphism from V to its dual? $\endgroup$
    – Salvatore
    Sep 11, 2016 at 12:38
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    $\begingroup$ @Salvatore Let $K$ be the underlying field and $f_1:V \rightarrow K$ be $f_1(z)=\varphi(x,z)$. $\endgroup$
    – Daniel
    Sep 11, 2016 at 12:43
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    $\begingroup$ @Salvatore $f_1,f_2$ are linear functionals from $V$ to the underlying field. The induced isomorphism is $x\rightarrow\varphi(x,\cdot)$. $\endgroup$
    – Daniel
    Sep 11, 2016 at 12:48
  • $\begingroup$ It is false. There is no basis of $\mathbb R^2$ such that $\begin{bmatrix}0&a\\a&0\end{bmatrix}$ represents the Euclidean dot product. $\endgroup$
    – mr_e_man
    May 16, 2023 at 21:15
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    $\begingroup$ @mr_e_man You are right. The result indeed fails for real vector spaces. However it is true if the field is algebraically closed. $\endgroup$
    – Daniel
    May 17, 2023 at 0:40

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