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I seek an example of a nonzero $\Bbb{R}$-bilinear map $f:V\times V\rightarrow W$ on a vector space $V$ (s.t: $\dim V<\infty$, $\dim W<\infty$) such that it is degenerate map, where $V$ and $W$ are given explicitly.

Recall that: A degenerate bilinear form $f:V\times V\rightarrow W$ on a finite-dimensional vector space $V$ is a bilinear form such that it has a non-trivial kernel, i.e., there exist some non-zero $u$ in $V$ s.t $f(u,v)=0 $ for all $v\in V$.

Thank you in advance

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  • $\begingroup$ But $\dim V<0$ does not make sense here. What about the zero map ? $\endgroup$ Jun 29, 2016 at 13:15
  • $\begingroup$ yeah, I'm sorry. I would write $\dim V<\infty$. $\endgroup$
    – Z. Alfata
    Jun 29, 2016 at 13:29
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    $\begingroup$ So take $f(u,v)=0$ for all $u,v$. This is a "highly" degenerated bilinear map. $\endgroup$ Jun 29, 2016 at 13:30
  • $\begingroup$ I seek an example of nonzero bilinear map $\endgroup$
    – Z. Alfata
    Jun 29, 2016 at 13:32
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    $\begingroup$ No problem. Just take the map which is "almost" zero, i.e., zero except for, say, $f(e_1,e_1)=f_1$, and $f(e_i,e_j)=0$ for all $i,j>1$, where $e_i$, $f_i$ are basis vectors of $V$ respectively $W$. $\endgroup$ Jun 29, 2016 at 13:41

2 Answers 2

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recall if $f/V\times V \rightarrow W$ is a bilinear map and $W=K$ the field of scalar, then $f$ is sayd a bilinear fom.

if $dimV=1$ there are no degenerate bilinear map othere the zero map, so all non zero bilinear map are non degenerat.

if $dimV>1$ the example above in comment is on, anothere tack $f(e_1,e_j)=0$ and somme one $f(e_i,e_j)\not=0$ for $i>1$ and $j\geq 1$.

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  • $\begingroup$ But, for $V=\mathbb R=W$ (so $\dimV=\dim W =1$) there is a non-degenerate bilinear map, it is $f(x,y)=\alpha xy$, where $\alpha \in \mathbb R$? $\endgroup$
    – Z. Alfata
    Jun 30, 2016 at 0:12
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    $\begingroup$ I apologize this is a typing error , forget one " no " thanks for the correction $\endgroup$
    – m.idaya
    Jun 30, 2016 at 0:18
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A bilinear form $f:V\times V \to \mathbb{K}$ takes values in the underlying field of scalars $\mathbb{K}$ of $V$, and is entirely determined by its action on pairs of basis elements of $V$. If $V$ is finite-dimensional, $\dim V = n$, then each choice of basis $(e_i)_{i=1}^n$ for $V$ induces an isomorphism between the space of bilinear forms and the space of $n\times n$ matrices with values in $\mathbb{K}$: the isomorphism maps the form $f$ to the matrix $(f(e_i,e_j))$, and its inverse maps the matrix $A$ to the form $((u,v) \mapsto u^t A v)$.

This yields as many examples of bilinear forms as there are $n \times n$ matrices. Moreover, under this isomorphism, degenerate forms correspond to singular matrices. So for $n = 2$, for instance, a form $f$ is degenerate iff $\det (f(e_i,e_j)) = f(e_1,e_1)f(e_2,e_2) - f(e_1,e_2)f(e_2,e_1) = 0$.

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