# The Unapologetic Mathematician

## Mathematics for the interested outsider

Since an inner product on a finite-dimensional vector space $V$ is a bilinear form, it provides two isomorphisms from $V$ to its dual $V^*$. And since an inner product is a symmetric bilinear form, these two isomorphisms are identical. But since duality is a (contravariant) functor, we have a dual transformation $T^*:W^*\rightarrow V^*$ for every linear transformation $T:V\rightarrow W$. So what happens when we put these two together?

Say we start with linear transformation $T:V\rightarrow W$. We’ll build up a transformation from $W$ to $V$ which we’ll call the “adjoint” to $T$. First we have the isomorphism $W\rightarrow W^*$. Then we follow this with the dual transformation $T^*:W^*\rightarrow V^*$. Finally, we use the isomorphism $V^*\rightarrow V$. We’ll write $T^*$ for this composite, and rely on context to tell us whether we mean the dual or the adjoint (but because of the isomorphisms they’re secretly the same thing).

So why is this the adjoint? Let’s say we have vectors $v\in V$ and $w\in W$. Then it turns out that

$\displaystyle\left\langle T(v),w\right\rangle_W=\left\langle v,T^*(w)\right\rangle_V$

which should recall the relation between two adjoint functors. An important difference here is that there is no distinction between left- and right-adjoint transformations. The adjoint of an adjoint is the original transformation back again: $\left(T^*\right)^*=T$. This follows if we use the symmetry of the inner products on the relation above

$\displaystyle\left\langle w,T(v)\right\rangle_W=\left\langle T^*(w),v\right\rangle_V=\left\langle w,T^{**}(v)\right\rangle_W$

Then since $\left\langle w,\left[T-T^{**}\right](v)\right\rangle_W=0$ and the inner product on $W$ is nondegenerate, we must have $T-T^{**}$ sending every $v$ to the zero vector in $W$. Thus $T=T^{**}$.

So let’s show this adjoint condition in the first place. On the left side, we have the result of applying the linear functional $\langle\underline{\hphantom{X}},w\rangle_W\in W^*$ to the vector $T(V)$. But this linear functional is simply the image of the vector $w$ under the isomorphism $W\rightarrow W^*$. So on the left, we’ve calculated the result of first applying $T$ to $v$, and then applying this linear functional.

But the way we defined the dual transformation was such that we can instead apply the dual $T^*$ to the linear functional $\langle\underline{\hphantom{X}},w\rangle_W$, and then apply the resulting functional to $v$, and we’ll get the same result. And the isomorphism $V^*\rightarrow V$ tells us that there is some vector in $v_1\in V$ so that the linear functional we’re now applying to $v$ is $\langle\underline{\hphantom{X}},v_1\rangle_V$. That is, our value will be $\langle v,v_1\rangle_V$ for some vector $v_1$. Which one? The one we defined as $T^*(w)$.

I’ll admit that it sometimes takes a little getting used to the way adjoints and duals are the same, and also the subtleties of how they’re distinct. But it sinks in soon enough.

May 22, 2009 Posted by | Algebra, Linear Algebra | 7 Comments

## Testing

If all goes according to plan, a link to this post should show up on my Twitter feed shortly. Thanks to twitterfeed.