# The Unapologetic Mathematician

## Transformations of Bilinear Forms

Sorry for the delay, but at least now everything is in Maryland. Now I just need a job and then I’ll have to move back out of my parents’ house again

Anyway, now we want to see how linear maps between vector spaces affect forms on those spaces. We’ve seen a hint when we talked about the category of inner product spaces: if we have a bilinear form $B$ on a space $W$ and a linear map $T:V\rightarrow W$, then we can “pull back” the form along the map. That is, we take two vectors in $V$, apply $T$ to both of them, and then stick them into the form on $W$.

When we write this out with our Dirac notation, we don’t really make much of a distinction between the vector spaces, relying on context to tell which vectors belong to which spaces, as well as which pairing we’re using. Luckily, when we write something sensible, it usually doesn’t matter how we parse things. Dirac notation is very robust! So, given vectors $w_1$ and $w_2$ in $W$ we have the form given by

$\displaystyle\langle w_1\rvert B\lvert w_2\rangle$

as we’ve seen over and over. But now let’s take two vectors $v_1$ and $v_2$ in $V$. First, we hit each of these vectors with $T$. The ket vector $\lvert v_2\rangle$ clearly becomes $T\lvert v_2\rangle$, while the bra vector $\langle v_1\rvert$ becomes $\langle v_1\rvert T^*$. Now using the form from before we get

$\displaystyle\langle v_1\rvert T^*BT\lvert v_2\rangle$

This looks like a form on $V$ that’s described by the linear transformation $T^*BT$. Just as a sanity check, we can verify that $T$ sends $V$ to $W$, $B$ then sends $W$ to itself, and the adjoint $T^*$ sends $W$ back to $V$. So this is indeed a transformation from $V$ to itself that can be used to represent a form.

As a particular case, we might consider an automorphism $S:V\rightarrow V$ and consider it as a change of basis. When we did this to a linear map $T$ we got a new linear map by conjugation $S^{-1}TS$, and we say that these two transformations are similar. In one sense, this is “the same” linear map, but described with different “coordinates” on the vector space. But when we apply it to a bilinear form we turn $B$ into $S^*BS$, and we say that these two are “congruent”. In the same sense as before, they describe “the same” form on $V$, but using different coordinates on the vector space.

Like similarity, congruence gives an action of $\mathrm{GL}(V)$ on the space of bilinear forms on $V$. Indeed, if $S$ and $T$ are both automorphisms, we can act on $B$ first by $S$ to get $S^*BS$, then then by $T$ to get $T^*S^*BST$. But this is the same as $\left(ST\right)^*B\left(ST\right)$, which is the action by the product $ST$.

July 24, 2009 Posted by | Algebra, Linear Algebra | 7 Comments