## Adjoints!, or: Why Aren’t These in My Crappy Linear Algebra Book? part 2.

### July 25, 2010

This part is going to be really exciting, no lies. In fact, we’re really just going to prove one or two theorems and that’ll be that. The reason for doing so is because one of these theorems is so elegant and beautiful that I want you to focus on it. Specifically, the fact that the matrix associated to an adjoint linear map is the conjugate transpose of the matrix associated to the original linear map. Best.

Okay, let’s just get right into it, because you *know* all the stuff you need to know in order to do this theorem. The only thing we want to remind you of is the fact that, supposing we have an orthonormal basis for our nontrivial finite dimensional vector space , we can write any element as

The proof of this is in one of the previous posts, but, you know, you can try to do it yourself if you want. Either way, let’s begin.

**Theorem**: Let and be nontrivial finite dimensional vector spaces and let be a linear map. Then is the conjugate transpose of .

Proof.Okay, so, first, for both of these spaces we have an orthonormal basis. For we’re going to call these and for we’re going to call these .In order to find the -th column of , we need to write out in terms of the ‘s. Luckily, because we have an orthonormal basis, we know what is! Specifically, we have this is equal to

and, therefore, our matrix entry in the -th column in the -th row will be . Make sure you see why this is — it’s really just looking at where everything goes from the equation we wrote above.

Now, let’s replace with and we derive the exact equation, except since we have that

and so in the -th row and -th column will be, similar to above,

Now, let’s do some cool manipulation with this one. Okay? Follow me here.

which means, in particular, that has entries which are tranpose and conjugate of . This last equation shows this. Make sure you understand

whythis last equation implies it. Actually, we could sum it up better by excluding all the middle parts and just sayingwhich makes it slightly clearer. .

This proof, in particular, is so magical that it hurts me to love it so much. It’s so simple, and, yet, gives us this really elegant tool for computing what look, at first, to be very strange things.

Now, for homework, I want you to prove something for me. I want you to prove the following:

**Theorem**: For and are nontrivial finite dimensional vector spaces and is a linear map, then the dimension of the range of is the same as the dimension of the range of . In other words, .

Proof. By you! You shouldn’t need much more than the theorem before this, and perhaps a bit of manipulation of the rank-nullity theorem.

The astute reader here will notice something: what if the matrix is real and symmetric about the diagonal? Then the adjoint and the original linear map are the same! Is this the only time that this will happen? Are there cool properties that these *self-adjoint linear maps* have? You bet your petticoat there are! The next post will detail these, and maybe more. Maybe. If you’re good.