# Show that $\widetilde{f}$ is a nondegenerate map?

Noted by $V, W$ two vectors spaces over the same field $K$ of finite dimensions. Let $f:V\times V\rightarrow W$ a degenerate map. I would show that $$\widetilde{f}:\widetilde{V}\times \widetilde{V}\rightarrow W$$ is a nondegenerate map, where $\widetilde{V}:= V/\ker{f}$, with $\ker{f}=\{x\in V: \, f(x,y)=0\quad \forall y\in V\}$ is the subspace kernel of $f$.

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. And it is nondegenerate iff it's subspace kernel is trivial.

• I think you did not make it clearly. Is $f$ symmetric bilinear? And you should also define $\widetilde{f}$ clearly and prove that it's well-defined first of all. – Fish Jul 1 '16 at 7:57

This is a good question to see what happens in passing to the quotient, but we must to avoids some simple error. as the deference between bilinear map and bilinear form. and for a bilinear map there are two type of kernel that is left and right, and this two type are equal in the case of symmetric bilinear map.

so in your context we speak on the left kernel of $f$ as you have wrote: $kerf=\{x\in V \mid f(x,y)=0\forall y\in V\}$ and this is a well subspace of $V$ so $\widetilde{V}=V/kerf$ is a quotient space, but what is $\widetilde{f}$? that is the problem of lifting $f$ via the space $\widetilde{V}=V/kerf \times\widetilde{V}=V/kerf$, this is the problem : when $\widetilde{f} (\widetilde{x},\widetilde{y})=f(x,y)$ has a sense for all $x,y\in V$? if you don't suppose $f$ symmetric there no warrant $\widetilde{f}$ to be well defined.

so we answered the question in the case all good, that is $f$ symmetric and so $\widetilde{f}$ is well defined in this case.

in this case we tack $\{e_1,...,e_m\}$ a basis of $kerf$ and we complete this basis with $\{e_{m+1},...,e_{m+n}\}$ for obtaining a basis $\{e_1,...,e_m, e_{m+1},...,e_{m+n}\}$ of $V$ then $\{\widetilde{e}_{m+1},...,\widetilde{e}_{m+n}\}$ is a basis of the quotient space $V/kerf$.

we prove that $\widetilde{f}$ is not degenerate bilinear map. let $x\in V$ such that $\widetilde{f}(\widetilde{x}, \widetilde{e}_j)=f(x,e_j)=0,\;\forall j\in\{m+1,...,m+n\}$, this applique that $f(x,e_i)=0,\;\forall i\in\{1,...,m+n\}$ (because for $i\in\{1,...,m\}$ we have $e_i\in kerf$), so $x\in kerf$ and $\widetilde{x}=0$ this prove that $\widetilde{f}$ not degenerate.

• note: you can suppose only leftkerf=rightkerf the all are correct; thanks – m.idaya Jul 1 '16 at 11:49
• Thank you very much m.idaya. Just, I would like to know, where intervenes the hypothesis "symmetric" in your proof? – Z. Alfata Jul 1 '16 at 14:23
• to you too: intervene in $\widetilde{f}$ to be well defined; because if $\widetilde{x}=\widetilde{u}$ and $\widetilde{y}=\widetilde{w}$ we must have $\widetilde{f}(\widetilde{x},\widetilde{y})=\widetilde{f}(\widetilde{u},\widetilde{w})$. but this is true for all $x,u,y,w\in V$ iff leftkerf=rightkerf. so when $f$ is symmetric this hold i.e leftkerf=rightkerf. – m.idaya Jul 1 '16 at 14:47
• sorry I can't correct the latex error in my comment: to read $\widetilde{f}(\widetilde{x},\widetilde{y})=\widetilde{f}(\widetilde{u},\widetilde{w})$. all is get in my latex compilation but here are problem – m.idaya Jul 1 '16 at 14:55
• l$eftkerf=\{x\in V\mid f(x,y)=0\forall y\in V\}$ and $rightkerf=\{y\in V\mid f(x,y)=0\forall x\in V\}$ – m.idaya Jul 2 '16 at 8:00