Let $V=V_0 \oplus V_1$ be a $\mathbb{Z_2}$ finite dimensional graded vector space of dimension $2n$. Let $x_1,..,x_n,y_1,...,y_n$ be a basis of $V$ such that the $x_i$ form a basis for $V_0$ and the $y'i$ form a basis for $V_1$.
Define a bilinear form $B$ on $V$ by $B(x_i,y_j)= \delta_{ij}$ and $B(x_i,x_j)=0=B(y_i,y_j)$ for all $i,j$.
Let $L_B(V)$ be the algebra of graded endomorphisms $V \to V$ which preserve $B$.
Why is $L_B(V)$ isomorphic to $\Lambda^2(V^*)$, the second power of the graded symmetric algebra on the dual of $V$, i.e. strictly quadratic polynomials? I am reading a proof in which they identify these two.
Since bilinear forms can be naturally identified with $T^2(V)^*$, and $V$ is finite dimensional (so $T^2(V)^* \cong T^2(V^*)$ as vector spaces), you can construct the natural map $$ \begin{align} \alpha: T^2(V^*) &\to L_B(V) \\ f_1 \otimes f_2 &\mapsto \big( (v_1, v_2) \mapsto f_1(v_1)f_2(v_2) \big) \end{align} $$ and verify that $\alpha$ descends to an isomorphism $\Lambda^2(V^*) \to L_B(V)$.