Let $b : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ be a bilinear form, and $\langle,\rangle$ be the standard inner product of the Euclidean space and $e_1,...e_n$ be the standard basis for the Euclidean space.

Then, for arbitrary $x$ and $y$ in $\mathbb{R}^n$, I want to construct a matrix $B$ such that $b(x,y)=\langle Bx,y\rangle$.

Here, is it correct that the matrix $B$ is defined to be the 'transpose' of matrix $B'$ whose component in the $i$th row and $j$th column is $b(e_i,e_j)$?

Because $B$ must be in the first argument of the inner product, I think it is necessary to take the transpose. What do you think?


1 Answer 1


Yes because note that the standard inner product in $\Bbb R^n$ can be expressed like so: $$\langle p, q \rangle = p^Tq$$ where $p^T$ is the transpose of $p$.

Hence $b(x, y) = \langle Bx, y \rangle = (Bx)^Ty = x^TB^Ty$.

So $b(e_i, e_j) = e_i^T B^T e_j$ and the operation $e_i^T B^T e_j$ selects the $i^\text{th}$ row of $B^T$ and the $j^\text{th}$ row of $B^T$.

Or in other words $b(e_i, e_j) = [B^T]_{i, j} = B_{j, i}$. In your notation the matrix $B^T$ is of course $B'$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .