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 \{e_{1}, \dots, e_{n}\} for our nontrivial finite dimensional vector space V, we can write any element v\in V as

v = \langle v, e_{1}\rangle e_{1} + \langle v, e_{2}\rangle e_{2} + \cdots + \langle v,e_{n}\rangle e_{n}

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 V and W be nontrivial finite dimensional vector spaces and let T:V\rightarrow W be a linear map.  Then M(T^{\ast}) is the conjugate transpose of M(T).

Proof. Okay, so, first, for both of these spaces we have an orthonormal basis.  For V we’re going to call these \{e_{1}, e_{2}, \dots, e_{n}\} and for W we’re going to call these \{f_{1}, f_{2}, \dots, f_{m}\}.

In order to find the k-th column of M(T), we need to write out T(e_{k}) in terms of the f_{i}‘s.  Luckily, because we have an orthonormal basis, we know what T(e_{k}) is!  Specifically, we have this is equal to

T(e_{k}) = \langle T(e_{k}), f_{1}\rangle f_{1} + \langle T(e_{k}), f_{2}\rangle f_{2} + \dots + \langle T(e_{k}), f_{m}\rangle f_{m}

and, therefore, our matrix entry in the k-th column in the r-th row will be \langle T(e_{k}),f_{r}\rangle.  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 T with T^{\ast} and we derive the exact equation, except since T^{\ast}: W\rightarrow V we have that

T^{\ast}(f_{k}) = \langle T^{\ast}(f_{k}), e_{1}\rangle e_{1} + \dots + \langle T^{\ast}(f_{k}), e_{n}\rangle e_{n}

and so in M(T^{\ast}) the r-th row and k-th column will be, similar to above,

\langle T^{\ast}(f_{k}), e_{r}\rangle

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

\langle T^{\ast}(f_{k}), e_{r}\rangle = \langle f_{k}, (T^{\ast})^{\ast}(e_{r})\rangle

= \langle f_{k}, T(e_{r})\rangle = \overline{\langle T(e_{r}), f_{k}\rangle}

which means, in particular, that M(T^{\ast}) has entries which are tranpose and conjugate of M(T).  This last equation shows this.  Make sure you understand why this last equation implies it.  Actually, we could sum it up better by excluding all the middle parts and just saying

\langle T^{\ast}(f_{k}), e_{r}\rangle = \overline{\langle T(e_{r}), f_{k}\rangle}

which makes it slightly clearer.  \Box.

 

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 V and W are nontrivial finite dimensional vector spaces and T:V\rightarrow W is a linear map, then the dimension of the range of T is the same as the dimension of the range of T^{\ast}.  In other words, dim(range(T)) = dim(range(T^{\ast})).

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.

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