# 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

1. How do you generally work? In spurts? Or over like hours at a time?

Comment by ramanujantao | June 27, 2009 | Reply

2. In spurts between teaching (more and more each year) and applying for another job (which I’ve had to do every year). Which amounts to “not much”.

Comment by John Armstrong | June 27, 2009 | Reply

3. [...] in Dirac Notation Now, armed with Dirac notation, we can come back and reconsider matrices and forms. For our background, we’ve got an inner product space. That is, a vector space , equipped [...]

Pingback by Matrices and Bilinear Forms on Inner Product Spaces in Dirac Notation « The Unapologetic Mathematician | July 8, 2009 | Reply

4. Formula does not parse on the second line; missing a curly bracket, I think.

Comment by Tommi Brander | November 16, 2009 | Reply

5. Weird. That wasn’t there when I posted it.

Comment by John Armstrong | November 17, 2009 | Reply