Throughout k is a field and all vetor spcaes are over k.

Definition 1. Let V be a k-vector space. A bilinear form on V is a map B : V \times V \longrightarrow k such that for every \alpha_1, \alpha_2 \in k and v_1,v_2,v \in V we have

1) B(\alpha_1v_1+\alpha_2v_2, v)=\alpha_1B(v_1,v)+ \alpha_2B(v_2,v),

2) B(v, \alpha_1v_1+\alpha_2v_2)=\alpha_1B(v,v_1)+\alpha_2B(v,v_2).

Definition 2. Let B be a bilinear form on a vector space V. we say that B is non-degenerate if B(v,w)=0, \ \forall v \in V, implies w=0. If \{v_1, \ldots , v_n \} is a basis for V, then the n \times n matrix whose (i,j)-entry is B(v_i,v_j) is called the matrix of B with respect to that basis of V.

Theorem. Let V be a finite dimensional vector space with a bilinear form B. Then B is non-degenerate if and only if  the matrix of B is invertible.

Proof.  Choose a basis \{v_1, \ldots , v_n \} for V. Let [B] be the matrix of B. Then [B] \bold{x}=\bold{0}, for some \bold{x} \in k^n, if and only if \sum_{j=1}^n B(v_i,v_j)x_j=0, for all 1 \leq i \leq n, where x_j is the j-th coordinate of \bold{x}. Thus [B] \bold{x}=\bold{0} if and only if B(v_i, \sum_{j=1}^n x_jv_j)=0, for all 1 \leq i \leq n, if and only if B(v, \sum_{j=1}^n x_j v_j)=0, for all v \in V. Hence [B] \bold{x}=\bold{0} has a non-zero solution for \bold{x} if and only if there exists 0 \neq w \in V such that B(v,w)=0 for all v \in V. Thus [B] is not invertible if and only if B is not non-degenerate. \Box

Remark. Since [B] is invertible if and only if [B]^T, the transpose of [B], is invertible, a similar argument to the above theorem gives us that B is non-degenerate if and only if B(w,v)=0, \ \forall v \in V, implies w=0.

Definition 3. A finite dimensional k-algebra A is called a Frobenius algebra if A, as a k-vector space, has a non-degenerate bilinear form B such that B(xy,z)=B(x,yz), for all x,y,z \in A.

Example. Let A=M_n(k), the algebra of n \times n matrices with entries in k. Then A is a Frobenius k-algebra.

Proof. Define B : A \times A \longrightarrow k by B(x,y)=\text{Tr}(xy), the trace of xy. It is clear that B is a bilinear form. To prove that B is non-degenerate, suppose that B(x,y)=0 for all x \in A. So \text{Tr}(xy)=0 for all x \in A. Let e_{ij} be the element of A with (i,j)-entry 1 and 0 anywhere else. Let y_{ij} be the (i,j)-entry of y. Let 1 \leq r,s \leq n and let x = e_{rs}. Then 0=\text{Tr}(xy)=y_{sr} and so y=0. This proves that B is non-degenerate. Finally, for any x,y,z \in A, we have B(xy,z)=\text{Tr}((xy)z)=\text{Tr}(x(yz))=B(x,yz). \ \Box

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