# Are non-degenerate bilinear forms equivalent to isomorphisms $V \cong V^*$?

$\newcommand{\Bilinear}{\operatorname{Bilinear}}$ Let $V$ be a finite-dimensional vector space over the field $\mathbb{F}$, and let $V^*$ denote its dual. Denote by $\mathscr{L}(V,V^*)$ by the space of all linear transformations from $V$ to $V^*$. I define a bilinear form to be a bilinear function $V \times V \to \mathbb{F}$. A bilinear form $\beta$ is said to be non-degenerate if for any $x\in V$, $\beta(x, \cdot) \equiv 0 \implies x =0$.

1. Are the spaces $\Bilinear(V\times V, \mathbb{F})$ and $\mathscr{L}(V,V^*)$ linearly isomorphic? In other words, does every bilinear form uniquely determine a homomorphism $V \to V^*$, and vice versa, in such a manner that respects the linear structure of both spaces?

2. Does every non-degenerate bilinear form uniquely determine an isomorphism $V \to V^*$, and does every isomorphism $V \to V^*$ uniquely determine a non-degenerate bilinear form, in such a way that respects the linear structure of both spaces?

Let me offer a short proof using dual bases. Let $v_1, \dots, v_n$ be some basis of $V$ and $v^1, \dots, v^n$ the associated dual basis of $V^{*}$. Define the map $\phi \colon \mathcal{L}(V,V^{*}) \rightarrow \operatorname{Bi}(V \times V, \mathbb{F})$ by $\phi(T) := g_T$ where $g_T(u,v) = T(u)(v) = \varepsilon(T(u), v)$. It is readily verified that this map is linear so it is enough to show that $\phi$ is one-to-one and onto. Since $$T(v_i) = \sum_{i=1}^n \varepsilon(T(v_i), v_j) v^j$$

we see that the map $T$ is determined uniquely by the $n \times n$ scalars $\varepsilon(T(v_i), v_j)$ (since the matrix $(\varepsilon(T(v_i),v_j))$ is the matrix representing $T$ with respect to the bases $v_1,\dots,v_n$ of $V$ and $v^1,\dots,v^n$ of $V^{*}$). Hence, if $\phi(T) = 0$ then all the scalars $\varepsilon(T(v_i), v_j)$ are zero and $T = 0$. This shows the injectivity.

To show surjectivity, let $g$ be a bilinear form on $V$ and define a linear map $T$ by requiring that

$$T(v_i) = \sum_{i=1}^n g(v_i, v_j)v^j$$

for all $1 \leq i \leq n$. Then a direct calculation shows that $\phi(T) = g$.

• This is helpful, since my next step was to figure out how to represent everything with matrices, and how to choose a dual basis in order to do so -- this already explains that, and quite concisely, so the argument can be followed along easily, but at the same time there is still room to work out details, making for good practice/learning. I appreciate it! – Chill2Macht Jan 16 '17 at 13:40
• @William: You're welcome! It's a good exercise to fill in all the details since these type of arguments repeat all over the place when working with multilinear maps/tensor products. – levap Jan 16 '17 at 13:43

1. The answer to this question is yes. The proof is analogous to that in this question.

Let $\varepsilon: V^* \times V \to \mathbb{F}$ be the evaluation map, which for all $f\in V^*$ and $v\ \in V$ is defined as: $$\varepsilon: (f,v) \mapsto f(v)\,.$$ $\varepsilon$ is clearly bilinear. Given a linear transformation $T \in \mathscr{L}(V,V^*)$, we define a bilinear form by: $$\Xi:T \mapsto \varepsilon(T(\cdot),\cdot) \,.$$ This function is linear, since given $s,t \in \mathbb{F}, S,T \in \mathscr{L}(V,V^*)$, by the bilinearity of $\varepsilon$: $$\varepsilon((sS+tT)(\cdot),\cdot )= \varepsilon(sS(\cdot)+tT(\cdot),\cdot)=s\varepsilon(S(\cdot),\cdot)+t\varepsilon(T(\cdot),\cdot) \,.$$ To conclude the proof, we show that $\Xi: \mathscr{L}(V,V^*) \to \Bilinear(V \times V, \mathbb{F})$ is bijective (1)(2).

Injectivity: This is essentially a computation using linearity. If you are unsure about any step please feel free to comment on this post. Assume that, for $S,T \in \mathscr{L}(V,V^*)$, $\Xi S = \Xi T$. Then: $$\Xi S = \Xi T \iff \varepsilon(S(\cdot),\cdot)=\varepsilon(T(\cdot),\cdot) \\ \iff 0=\varepsilon(S(\cdot),\cdot)-\varepsilon(T(\cdot),\cdot) =\varepsilon(S(\cdot)-T(\cdot),\cdot)=\varepsilon((S-T)(\cdot),\cdot) \,.$$

For any fixed $f\in V^*$, clearly, $\varepsilon(f, v)=f(v)=0$ for all $v \in V$ if and only if $f = 0$.

By the above, for any fixed $f \in V^*$, $$\varepsilon((S-T)f, v)=0$$ for all $v \in V$. It follows that $(S-T)f=0$ for any $f \in V^*$. Thus $S-T=0$ and $S=T$. $\square$

Now, if one knows already that $\mathscr{L}(V,V^*)$ and $\Bilinear(V\times V, \mathbb{F})$ have the same dimension, then the above actually suffices to show that $\Xi$ is bijective.

However, a direct proof of surjectivity of $\Xi$ not only shines light onto why this proposed isomorphism works, but is also no more difficult than a verification of the fact that the domain and target of $\Xi$ have the same dimension. Thus we will use this approach here.

Surjectivity: Let $\beta \in \Bilinear(V \times V, \mathbb{F})$. Fix a basis $v^1, \dots, v^n$ of $V$. Then for all $i =1,\dots,n$, $$\beta(v^i,\cdot ) \in V^* \,.$$ Using the fact that any linear transformation is determined by its action on a basis, we define $T \in \mathscr{L}(V,V^*)$ by $\quad Tv^i = \beta(v^i, \cdot) \quad$ for all $i$.

Claim: $\beta = \varepsilon(T(\cdot),\cdot)$. Let $u=u_1v^1+\dots+u_nv^n, w \in V$ be arbitrary. Then: $$\varepsilon(Tu,w)=\varepsilon(T(u_1v^1+\dots+u_nv^n),w)=u_1\varepsilon(Tv^1,w)+ \dots + u_n\varepsilon(Tv^n,w) \\ = u_1 \varepsilon(\beta(v^1,\cdot),w)+\dots+u_n\varepsilon(\beta(v^n,\cdot),w) = u_1 \beta(v^1,w)+\dots+u_n \beta(v^n,w) \\ = \beta(u_1v^1+\dots+u_nv^n,w)=\beta(u,w)\,. \square$$

2. Using the first part, all we need to show is that: $$T \in \mathscr{L}(V,V^*)\text{ is invertible} \iff \Xi T =\beta \in \Bilinear(V \times V, \mathbb{F})\text{ is non-degenerate.}$$ Two observations make this easier to show.

(i) Since $\dim V = \dim V^*$, $T$ is invertible $\iff$ $T$ is injective.

(ii) For any $x \in V$ (i.e. not just $v^1, \dots, v^n$), $Tv=\beta(x,\cdot)$.

Proof of (ii): Let $x=x_1v^1+\dots+x_nv^n$. Then: $$Tx=T(x_1v^1+\dots+x_nv^n) =x_1Tv^1+\dots+x_nTv^n= x_1 \beta(v^1,\cdot)+\dots+x_n\beta(v^n,\cdot)=\beta(x_1 v^1+ \dots+x_nv^n,\cdot)=\beta(x,\cdot)\,.$$

$\beta$ is non-degenerate $\implies$ $T$ is injective: Assume $Tx_1 = Tx_2$. Then: $$0=Tx_1 - Tx_2 = T(x_1-x_2)=\beta(x_1-x_2,\cdot)=0 \implies x_1 - x_2=0 \iff x_1=x_2\,,$$ the implication following from the non-degeneracy of $\beta$. Thus $T$ is injective. $\square$

$T$ is injective $\implies$ $\beta$ is non-degenerate: Because $T$ is injective, $Tx=T(0)=0 \implies x=0$. Since $Tx = \beta(x,\cdot)$, this means that $\beta(x,\cdot)=0 \implies x=0$. $\square$

• Note that there are actually two "natural" isomorphisms, given by $T \mapsto g_T(v,w) := T(v)(w)$ and $T \mapsto h_T(v,w) := T(w)v$ and they are not the same. – levap Jan 16 '17 at 12:22
• @levap Wouldn't the other one of the two give a bilinear form $V^* \times V^* \to \mathbb{F}$ though? Or is that what you mean by "they are not the same"? – Chill2Macht Jan 16 '17 at 12:34
• They both give you bilinear forms on $V \times V \rightarrow \mathbb{F}$. The point is that the space $\operatorname{Bilinear}(V \times V, \mathbb{F})$ has a natural automorphism which sends a bilinear form $(u,v) \mapsto g(u,v)$ to the bilinear form $(u,v) \mapsto g(v,u)$ (the same form, only we change the order of the arguments). The two possible isomorphisms of $\mathcal{L}(V,V^{*})$ to $\operatorname{Bilinear}(V \times V, \mathbb{F})$ are related to each other by this automorphism. – levap Jan 16 '17 at 12:40