# The Unapologetic Mathematician

## Matrices and Forms II

Let’s pick up our discussion of matrices and forms and try to tie both views of our matrix together. We’re considering a bilinear form $B:V\otimes V\rightarrow\mathbb{F}$ on a vector space $V$ over the real or complex numbers, which we can also think of as a linear map from $V$ to its dual space $V^*$. We can also consider a sesquilinear form on a complex vector space, which is equivalent to an antilinear (conjugates scalars) map from the space to its dual.

We got a matrix by picking a basis $\left\{e_i\right\}$ for $V$ and plugging basis vectors into our form $\displaystyle b_{ij}=B(e_i,e_j)$

And we also found that this is the matrix of the map from $V$ to $V^*$, written in terms of the basis $\left\{e_i\right\}$ and its dual basis $\left\{\epsilon^i\right\}$.

Now we have two bases of the same cardinality, so there is a (highly non-canonical) linear isomorphism from $V^*$ to $V$ which sends the basis vector $\epsilon^i$ to the basis vector $e_i$. If we compose this with the map from $V$ to $V^*$ given by the form $B$, we get a linear map from $V$ to itself, which we will also call $B$. If $B$ is sesquilinear, we use the unique antilinear isomorphism sending $\epsilon^i$ to $e_i$, and again get a linear map $B:V\rightarrow V$. The matrix of this linear map with respect to the basis $\left\{e_i\right\}$ is $b_i{}^j=b_{ij}$, just as before.

This seems sort of artificial, but there’s method here. Remember that we can come up with an inner product $\langle\underline{\hphantom{X}},\underline{\hphantom{X}}\rangle$ by simply declaring the basis $\left\{e_i\right\}$ to be orthonormal. Then the linear functional $\epsilon^i$ is given by $\langle e_i,\underline{\hphantom{X}}\rangle$. It’s this correspondence that is captured by both the non-canonical choice of inner product, and by the non-canonical choice of isomorphism.

But now we can extract the same matrix in a slightly different way. We start by hitting the basis vector $e_j$ with the new linear transformation $B$ to come up with a linear combination of the basis vectors. We then extract the appropriate component by using the inner product. In the end, we find $\displaystyle\langle e_i,B(e_j)\rangle=b_{ij}=B(e_i,e_j)$

So in the presence of a given basis (or at least a given inner product) any bilinear (or sesquilinear) form corresponds to a linear transformation, and the matrix of the linear transformation with respect to the selected basis is exactly that of the form itself. On the other hand, if we have a linear transformation from $V$ to itself, we can stick it into the inner product as above and get a bilinear (or sesquilinear) form out.

June 25, 2009 Posted by | Algebra, Linear Algebra | 5 Comments