# Normal Operator Equality

1. Nov 26, 2013

### TheOldHag

I'm having trouble seeing how this equality is possible which is seen in proofs of properties of normal operators.

||Tv||^2 = <T*Tv, v> = <TT*v, v> = ||T*v||^2

As far as I can get is

||Tv||^2 = <Tv, Tv>

2. Nov 26, 2013

### TheOldHag

I have made it an extra step.

||Tv||^2 = <Tv, Tv> = <v, T*Tv> = <v, TT*v> = ||T*v||^2

I got this from the definition of adjoint that I was overlooking because I was not thinking of Tv in the second slot as an arbitrary vector for some reason. Nonetheless, I'm stuck with the expression in the second slot. I can see why the ends of the equality are equal now but not sure how the examples have the expressions they get in the first slots of the inner products. These are complex vector spaces so there is a distinction.

Perhaps the issue is that the book I'm working with "Linear Algebra Done Right" defines the adjoint starting with a linear functional that maps to <Tv, v> and then proceeds to <v, T*v> based on a uniqueness theorem and never shows symmetrical properties of this process. Hence I'm a bit confused.

3. Nov 26, 2013

### AlephZero

If it helps, it is easy to see this if you represent T as a matrix and v as a vector. All the expressions are
$$\sum_i\sum_j\sum_k t_{ij} t^*_{ik} v_i v^*_k$$

Of course that "proof" might not count as "linear algebra done right", but if it gives some clues it's still useful.

4. Nov 26, 2013

### TheOldHag

I believe I figured out the above by using the property t** = t.

However, with this symmetry the next question is, why is the adjoint defined starting with <Tv, v> rather than <v, Tv>. It is the case that <Tv, v> = <v, T*v> = <v, Tv> if T is self adjoint only. But then why is there no mention of a left adjoint and a right adjoint.

The adjoint was initially not intuitive for me to grasp via the 'Done Right' book and I think it would have helped if it was noted that the most natural way to describe the action of a linear operator on a inner product space as a real number for each vector in the space would be to take <Tv, v> which combined v, Tv, and the notion of inner product into one real quantity. It does mention the analogy between adjoints and complex numbers since <Tv, v> = complex conjugate of <Tv, v>.

Sorry, rambling a bit. Thinking out loud regarding the adjoint and trying to grasp it intuitively.

5. Nov 26, 2013

### PeroK

If you pick a basis for the Vector space and express any two vectors in that basis, it is not hard to show from the properties of an inner product that:

For all T, u and v <Tu.v> = <u.T*v>

In fact, for a general Hilbert Space, that's the way the adjoint operator is defined.

So, for a normal operator, you have:

$||Tv||^2 = <Tv.Tv> = <v.T^*Tv> = <v.TT^*v> = <T^*v.T^*v> = ||T^*v||^2$

6. Nov 26, 2013

### Office_Shredder

Staff Emeritus
The reason there's no left adjoint and right adjoint is that they are the same. Suppose that I have my operation T, and there exists some S such that
<Tu,v> = <u,Sv>.

Then
$$\left< u,Tv \right> = \overline{ \left<Tv, u \right>} = \overline{ \left<v, Su \right> } = \left<Su,v \right>$$
so if S is a right adjoint, it's a left adjoint, and vice versa.

7. Nov 26, 2013

### TheOldHag

This kind of solidifies for me why the adjoint is so fundamental. It is the most intuitive way to summarize the action of an operator for each vector in an inner product space as a real number that includes v, Tv, and the inner product of both. And as you have pointed out, it is unique and not one-sided. Thanks.

8. Nov 26, 2013

### TheOldHag

Actually, I mean to say, the linear functional that leads to the definition of an adjoint <Tv, v> is fundamental. When I first started leaning this last week the adjoint seemed a bit out there.