# Adjoint of a linear operator

1. Dec 7, 2007

### kingwinner

Definition from my textbook: For each linear operator T on a inner product space V, the adjoint of T is the mapping T* of V into V that is defined by the equation <T*(v),w> = <v,T(w)> for all v, w E V.

My instructor defined it by <T(v),w> = <v,T*(w)> and he said that these 2 definitions are equivalent.

Now, can someone please explain WHY they are equaivalent?

Can both definitions be used at the SAME time, or do I have to choose 1 of the 2 definitions and use this chosen definition consistently everywhere?

Thanks for explaining!

2. Dec 7, 2007

### ozymandias

Hey kingwinner,

They are equivalent because you can prove that the adjoint of the adjoint is the operator itself.

Here's a longer explanation: suppose one definition tells us that
<T*(v),w> = <v,T(w)>
and the other that
<T(v),w>=<v,T@(w)>
Note that I've used different symbols (*,@) to denote conjugation, because we are not yet sure it is the same thing.
It is possible to show, using these definitions, that
T@@ = T
and
T** = T
(I won't go into the details here.)

Assuming this, we have, on the one hand:
<T(v),w> = <T**(v),w> = <(T*)*(v),w> = <v,T*(w)>
and on the other (by definition!):
<T(v),w> = <v,T@(w)>
Since this is true for all v,w, we must have T@ = T*.

Hope this helps :).

--------
Assaf
http://www.physicallyincorrect.com/" [Broken]

Last edited by a moderator: May 3, 2017
3. Dec 7, 2007

### HallsofIvy

Staff Emeritus
"Adjoint" is a dual property: If A* is the adjoint of A, the A is the adjoint of A* so either way works.

4. Dec 7, 2007

### Chris Hillman

Isn't it obvious? The inner product operation is symmetric, and so is the equality relationship.

5. Dec 7, 2007

### kingwinner

Thanks!

So if I am solving a problem, can both definitions be used at the SAME time within the same problem, or do I have to choose 1 of the 2 definitions and use this chosen definition consistently everywhere?

6. Dec 7, 2007

### ozymandias

Since the definitions are equivalent, you may use either at will.

--------
Assaf
http://www.physicallyincorrect.com/" [Broken]

Last edited by a moderator: May 3, 2017
7. Dec 8, 2007

### HallsofIvy

Staff Emeritus
Since T* is also linear, T*(u,v) must be (au+ bv, cu+ dv) for some complex numbers, a, b, c, and d.

You want $<(x+(1-i)y, (1+i)x+2y), (u,v)>= xu+ (1-i)\overline{u}y+ (1+i)x\overline{v}+ 2y\overline{v}$ to be equal to
$< (x,y), (au+ bv,cu+dv)>= \overline{a}x\overline{u}+ \overline{b}x\overline{v}+ \overline{c}\overline{u}y+ \overline{d}\overline{v}y$. Since this must be true for all x, y, u, v, coefficients of the same products must be equal: $1= \overline{a}, 1- i= \overline{c}, 1+i= \overline{b}, 2= \overline{d}$. That is, z= 1, b= 1- i, c= 1+i, and d= 2. T*(x,y)= (x+ (1-i)y, (1+i)x+ 2y). That happens to be exactly the same as T! This T is "self-adjoint".

I did that directly from the definition of adjoint to show how it works.
It is much simpler to just write this as a matrix:
$$T= \left[\begin{array}{cc}1 & 1- i \\ 1+i & 2\end{array}\right]$$
and take the "Hermitian adjoint": first swap row and columns (the "transpose")
$$T= \left[\begin{array}{cc}1 & 1+ i \\ 1-i & 2\end{array}\right]$$
then take the complex conjugate of each number.
$$T= \left[\begin{array}{cc}1 & 1- i \\ 1+i & 2\end{array}\right]$$
again, we see that T is self adjoint.

For linear transformations on a finite dimensional vector space over the real numbers, where the transformation can be written as a matrix with real number entries, its "adjoint" is the transpose and a linear transformation is "self adjoint" if and only if its matrix form is symmetric.

For linear transformations on a finite dimensional vector space over the complex numbers, where the transformation can be written as a matrix with complex number entries, its "adjoint" is the "Hermitian adjoint" described above. A transformation is "self adjoint" if and only if its matrix form is "Hermitian"- amn is the complex conjugate of anm.

By the way, if a linear transformation is from a vector space U of dimension m to a vector space V of dimension n, then it can be written as an m by n matrix. Its adjoint is a linear transformation from V to U and can be written as an n by m matrix- again the transpose.

Of course, to be "self adjoint", a matrix must be from vector space U to itself and represented by square matrix.

Last edited: Dec 8, 2007
8. Dec 8, 2007

### kingwinner

Thanks a lot!

But what is the relation between (u,v) and (x,y)? The question is looking for T*(x,y), but in your calculations, it seems that (u,v) is used instead of (x,y). I get a little confused here...

9. Dec 9, 2007

### CompuChip

You have two points in C^2. We can specify each point by giving two complex coordinates (e.g. (1, i) or (3+6i, 2-4i)). HallsOfIvy is taking two arbitrary points, and calls the coordinates of one of them x and y, and the coordinates of the other u and v.
Of course, these names are arbitrary. So if the calculation in the end shows that
T*(u,v) = (au+ bv, cu+ dv)
then obviously
T*(x,y) = (ax + by, cx+dy)
since x, y, u and v are just dummy placeholders (variables) for which you can choose any name. But since you are dealing with two different points (one in each slot of the inner product) you need to have at least four coordinates.

10. Dec 9, 2007

### HallsofIvy

Staff Emeritus
Surely you understand that if I calculate that f(u)= u2, I can also write f(x)= x2? It's the same thing.

If T is a linear transformation from inner product space V to inner product space U, then its adjoint is a linear transformation from U to V. If v is any vector in V and u is any vector in U, then the condition for an adjoint is <Tu, v>= <u, T*v> where the left side is the inner product in U and the right side is the inner product in V.

Your example was a linear transformation from V= C2 to U= C2. You had already defined T in terms of x and y so I took (x,y) to be my vector in V. In order NOT to confuse things by using the same variables to mean something else, I took (u, v) to be my vector in U.

11. Dec 9, 2007

### kingwinner

Thanks!

I have one more question about inner product.

Let u=(u1,u2), v=(v1,v2)
I denote (u1)* as conjugate of u1

How can I prove that <u,v>=2(u1)*v1 + (1-i)(u1)*v2 + (1+i)(u2)*v1 + 3(u2)*v2 defines an inner product on C2?

Now
[2 1-i
1+i 3]
is a Hermitian matrix, so if I can prove that <v,v> >0 and <v,v>=0 iff v=0, then I'm done, but how can I do so?

12. Dec 10, 2007

### CompuChip

To prove that it is an inner product, you should check that it satisfies the properties of an inner product.

For the question on the last line, you just plug in u1 = v1, u2 = v2 and work out the equation. You will notice that the i's drop out and you can write the result in the form
(... + ...)^2 + ...^2 + ...^2
from which it is easy to complete the proof.

13. Dec 10, 2007

### kingwinner

<v,v>=2(v1)*v1 + (1-i)(v1)*v2 + (1+i)(v2)*v1 + 3(v2)*v2

(v1)*v2 and (v2)*v1 won't cancel, so how can the i's cancel?

14. Dec 11, 2007

### CompuChip

It's not necessary that the entire terms cancel, just the stuff with an i in it
If I work out the brackets on the second and third term, I get
v1 * v2 - i * v1 * v2 + v2 * v1 + i * v2 * v1
And since v1 and v2 are just components of a vector, they are numbers and commute; v1 * v2 = v2 * v1

15. Dec 11, 2007

### kingwinner

Let v1=1+i, v2=1
v1* v2=(1-i)(1)=1-i
v2* v1= 1 (1+i) = 1+i

So in this example, they don't commute...

16. Dec 11, 2007

### CompuChip

Ooohh, confusion over the star
I used it as a multiplication sign, you used it as the conjugate.
Appearantly it still works out then, but not as easily as I thought... probably it is going to involve splitting everything in real and imaginary parts... I'll see if I can find an easier way.