In my linear algebra class today, the following calculation about bilinear forms was performed:
Let $b$ be a bilinear form on $V$ and $x,y \in V$. Suppose also $\{v_1,...,v_n\}$ is a basis for V. Then we can write: $$b(x,y) = b(x^iv_i, y^jv_j)\\ = x^iy^jb(v_i,v_j)\\$$ Then if you let $M_{ij} = b(v_i,v_j)$, we can write this as: $$b(x,y) = x^i M_{ij}y^j$$
I don't get the last step, intuitively I think it should be $$b(x,y) = x^iy^jM_{ij}$$$. Are the two formulas equivalent? Also why are we allowed to do that?
Since $x$ and $y$ are in $V$ they can be written as linear combinations of the basis vectors s.t $x=\sum_{i=1}^nx^iv_i$ and analogously for $y$ you can write $y=\sum_{j=1}^ny^jv_j$.($x^i , y^j$ are scalars and $v_i,v_j$ are basisvectors). Plugging this to formulas in the bilienar form you get $$b(x,y)=b(\sum_{i=1}^nx^iv_i,\sum_{j=1}^ny^jv_j)=\sum_{i=1}^n\sum_{j=1}^nx^iy^jb(v_i,v_j)$$where in the last steps bilinearity is used a couple of times. With the representation of $b(v_i,v_j)$ as a matrix $M_{i,j}$ you get the desired equality $$\sum_{i=1}^n\sum_{j=1}^nx^iy^jM_{i,j}$$ .Furthermore you can interchange the multiplication since the multplication of two scalaras and a scalar with a matrix is commutative.