## Adjoints!, or: Why Aren’t These in My Crappy Linear Algebra Book? part 1.

### July 24, 2010

Now, for this post, we’re going to assume something kind of unique: namely, we’re going to assume you know how to work inner products.  Yes, alas, I’m going to make a leap of faith here — but, reader, do not let me down!  I’ve posted a pdf explaining what they are a few posts ago, and we’ll be going over some basic properties of them as I go on in this post.  But the best way to learn these is to really do a bunch of problems that deal with them.  Almost every linear algebra book that I’ve seen has a huge section on these.

On the other hand, nearly every (basic) linear algebra book that I’ve seen has either a passing mention of adjoints or no mention at all.  This is more of an advanced topic, but it really doesn’t need to be — it’s not difficult, it’s just not all that intuitive.

But let’s stop talking about it and let’s start doing it.

Let’s let $V$ and $W$ be nontrivial finite dimensional vector spaces of dimension $n$ and $m$ respectively.  Let’s let $T: V\rightarrow W$.  Then we’re going to define the adjoint of $T$, which we write as $T^{\ast}$, in the following way:

• Fix an element $w\in W$ (which is arbitrary but remains fixed).
• Define $T^{\ast}$ by $\langle T(v), w\rangle = \langle v, T^{\ast}(w)\rangle$ for all $v\in V$.

And that’s it.  It looks much more simple than it actually is in most cases.  Let’s just do two easy examples so that you get the hang of it.

## Examples.

(1) Alright, let’s do a trivial example first.  If we have $V$ is a finite nontrivial vector space, then define $T:V\rightarrow V$ by $T(v) = v$.  Note that this is not a linear functional as we defined in the last post, because a linear functional goes from a vector space to its underlying field.  This particular linear map is called the identity map, and it goes from the space to itself.

So, let’s find the adjoint.  Let’s fix a $w\in V$ and then consider

$\langle T(v), w\rangle = \langle v, w\rangle = \langle v, T^{\ast}(w)\rangle$

What should we define $T^{\ast}$ as?  Well, looking at the last equality, it’s obvious: we should define $T^{\ast}(w) = w$ for all $w\in V$.  This way, the equality works.  Nice.  So the adjoint is also the identity map.  Weird.

(2) Let’s do a slightly less trivial example now.  Let’s let $V = {\mathbb R}^{3}$ and let’s consider the standard inner product on ${\mathbb R}^{3}$, and let’s let our linear transform be $T:{\mathbb R}^{3}\rightarrow {\mathbb R}$ defined by $T((x,y,z)) = x + 2y - z$.  Note that the number we get out is a scalar, so it lives in ${\mathbb R}$ as required!  Nice.  You can check that this is a linear map, or you can trust me; either way, let’s find the adjoint.  Let’s fix $w\in {\mathbb R}$.

$\langle T((x,y,z)), w\rangle = \langle x + 2y - z, w\rangle = wx + 2wy - wz$

and if we define $T^{\ast}(w) = (a,b,c)$ for some $(a,b,c)\in {\mathbb R}^{3}$, then

$\langle (x,y,z),T^{\ast}(w) \rangle = \langle (x,y,z),(a,b,c)\rangle = xa + yb + zc$

and making these two lines equal, we get that

$xa + yb + zc = wx + 2wy - wz$

which means that $a = w, b = 2w, c = -w$.  So we should define $T^{\ast}(w) = (w,2w,-w)$.  This gives us our adjoint.

You should’ve noticed by now that the adjoint is a little bit different from the regular linear map.  Also, if the linear map $T:V\rightarrow W$, then the adjoint will go from $T^{\ast}:W\rightarrow V$.  They swap domain and co-domain.

There are a whole bunch of properties that the adjoint and the regular linear map share, and we’ll prove them as we come to them.  But there are three properties which I really like proving, so we’re going to do them now.

Theorem: For $V, W$ nontrivial finite dimensional vector spaces and $F$ is a field (like the reals or the complexes), then we have for all $a\in F$ and $T:V\rightarrow W$ that $(aT)^{\ast} = \bar{a}T^{\ast}$.

Proof. By definition, we have

$\langle aT(x), y \rangle = a\langle T(x),y \rangle = a\langle x, T^{\ast}(y) \rangle = \langle x, \bar{a}T^{\ast}(y)\rangle$

which completes the proof.  $\Box$.

Theorem: For $V,W$ same as above, and $T:V\rightarrow W$, we have that $(T^{\ast})^{\ast}) = T$.

Proof. We have $\langle T(x),y \rangle = \langle x, T^{\ast}(y) \rangle = \langle (T^{\ast})^{\ast}(x), y\rangle$ and so the equality of the left-most-side and the right-most-side prove the theorem.  Make sure you know where the last equality is coming from.  $\Diamond$.

Theorem: If $V$ is the same as above, and $Id: V\rightarrow V$ is the identity operator, then $Id = (Id)^{\ast}$.  In other words, the identity is its own adjoint.

Proof. We have that

$\langle Id(x), y\rangle = \langle x,y \rangle = \langle x, Id^{\ast}(y)\rangle$

which implies that $Id^{\ast}(y) = y$ for all $y$, and so it is the identity function.  $\Diamond$.

One important theorem that I leave to you is the fact that the adjoint works in products the same kind of way that inverses do — namely, that

Theorem: For $U, V, W$ are nontrivial finite vector spaces with $T:V\rightarrow W$ and $S:W\rightarrow U$, then $(ST)^{\ast} = T^{\ast}S^{\ast}$.

Proof. Up to you!

I hope these theorems have been kind of fun to prove.  Either way, at this point, the adjoint should be a mysterious not-very-useful-looking thing, but we’re going to show in the next post that we have a very nice identity between $M(T)$ and $M(T^{\ast})$; namely, that they’re conjugate transposes of each other, which I’ll define next time.  This makes it so that we don’t have to go through a complicated calculation like above in the second example in order to find the adjoint of some map; if we have the matrix associated to it, it’s a near-trivial one-line manipulation to find it!  And that’s pretty sweet, no lies.