# Homework Help: Inner products and adjoints

1. Apr 6, 2013

### Zondrina

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

So I know that $T^*$ will be from F to V and we have a fixed scalar a in F. So :

$<u, T^*(a)> \space = \space <T(u), a> \space = \space <<u,v>,a>$

Now, I am unsure how to continue this. What should I do with <u,v>? I know that <u,v> is in F, but that doesn't seem to help me much?

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2. Apr 6, 2013

### Zondrina

Sorry for the double, but I stared at this for a second and thought of something.

$<<u,v>, a> \\ = < u_1 \overline{v_1} + ... + u_n \overline{v_n}, a> \\ = < u_1 \overline{v_1}, a> + ... + < u_n \overline{v_n}, a>$

I'm not quite sure this helps either though.

3. Apr 6, 2013

### micromass

You know that $<u,v>$ and $a$ are scalars. So what is $<<u,v>,a>$? What is the inner product in $\mathbb{R}$?

4. Apr 6, 2013

### Zondrina

Hmm... $<<u,v>,a> = a<<u,v>,1>\space$

its just a real number when we work over $ℝ$ I believe.

Last edited: Apr 6, 2013
5. Apr 6, 2013

### Dick

I don't get this whole question. V is a vector space. It doesn't say that it is F or R. If T maps V to F. Then T* doesn't map F to V. It maps the dual space V* to F. It doesn't act on scalars.

I'm confused. Or is there some interpretation of adjoint I don't know.

6. Apr 7, 2013

### Fredrik

Staff Emeritus
Suppose that H and K are Hilbert spaces over F, where F is either ℝ or ℂ. I will use the convention that when F=ℂ, the inner product is linear in the second variable. Suppose that $T\to K$ is linear and bounded. Let $x\in K$ be arbitrary. Define $\phi_x\to F$ by $\phi_x(y)=\langle x,Ty\rangle_K$ for all $y\in H$. This map is linear and bounded, so $\phi_x\in H^*$. The Riesz representation theorem for Hilbert spaces tells us that this implies that there's a unique $z\in H$ such that $\phi_x=\langle z,\cdot\rangle_H$. (The right-hand side denotes the map that takes an arbitrary $w\in H$ to $\langle z,w\rangle_H$). The map $x\mapsto z$ is denoted by $T^*$ and is called the adjoint of T. Note that $x\in K$ and $z\in H$, so $T^*:K\to H$.

Since $\phi_x=\langle T^*x,\cdot\rangle_H$, we have
$$\langle T^*x,y\rangle_H =\phi_x(y) =\langle x,Ty\rangle_K$$ for all $y\in H$.

The field of scalars F can be interpreted as a Hilbert space over F, with addition being the field's addition operation, and both scalar multiplication and the inner product being the field's multiplication operation. So the definition of "adjoint" above implies that if $T:V\to F$, then $T^*:F\to V$.

Last edited: Apr 7, 2013
7. Apr 7, 2013

### Zondrina

This is from the book linear algebra done right. So throughout the course of the chapter, V denotes a finite dimensional inner product space over F.

I was confused about it acting on scalars as well.

8. Apr 7, 2013

### Dick

Certainly. The only problem I have with that is that the problem doesn't say that V is F regarded as a vector space over itself. Unless this is in the context of other problems which say it is. And it's pretty sloppy to say $T^*:F\to V$ in that case. V and F have the same elements but V is F regarded as a vector space and F is a field. To say $T^*:F\to F$ is less sloppy. I don't see why you would want to reverse the symbols F and V.

Last edited: Apr 7, 2013
9. Apr 7, 2013

### Fredrik

Staff Emeritus
V can be any Hilbert space over F in this problem. It doesn't have to be equal to F.

10. Apr 7, 2013

### Zondrina

I don't believe we've covered Hilbert Spaces yet, so I don't think I can use anything pertaining to it.

V is an inner product space over a field F, but I don't think I've ever seen an inner product for scalars only which is why I'm having so much trouble manipulating the equation in my first post.

11. Apr 7, 2013

### micromass

The space $\mathbb{C}$ is always an inner-product space over itself. The inner product is given by $<x,y> = \overline{x}y$. You should check that this is indeed an inner product.
What will the inner product be of the space $\mathbb{R}$ over itself?

12. Apr 7, 2013

### Zondrina

It's the same as the complex one, except complex conjugation is not needed.

<x,y> = xy

13. Apr 7, 2013

### micromass

Right! So the product in $\mathbb{R}$ is a special case of the inproduct in an inner-product space! It's good to keep this in mind!

Can you solve the exercise now?

14. Apr 7, 2013

### Zondrina

$<u, T^*(a)> \space = \space <T(u), a> \space = \space <<u,v>,a> = \space <u \overline{v},a> = u \overline{va}$

Still can't see where this is going.

15. Apr 7, 2013

### micromass

No!! You can't say $<u,v> = \overline{u}v$!! That is only true in the special case $V=\mathbb{C}$. But your $u$ and $v$ are in a general space, so you can't use that expression.

The only thing you can conclude is

$$<u,T*(a)> = \overline{a}<u,v>$$

Now, can you use this to write $T*(a)$ in another form?? That is, do you know some other vector $x$ such that

$$<u,x> = \overline{a} <u,v>$$

16. Apr 7, 2013

### Fredrik

Staff Emeritus
Every finite-dimensional inner product space is a Hilbert space, so everything I said holds for this V as well. One thing is easier when we're dealing with a finite-dimensional space: We don't have to use the Riesz representation theorem.

Note however that what I wrote was an explanation meant for Dick. It may be useful for you too, but I was just talking about the definition of the adjoint in general. I haven't said anything about this specific problem.

17. Apr 7, 2013

### Zondrina

I just noticed my first post was wrong by a bit. I'll start over here and see what happens.

Okay so $T : V → F \Rightarrow T^*: F → V$. Fix a scalar $a \in F$ and we know $v \in V$. Also $T(u) = <u,v>$.

$<v, T^*(a)> \space = \space <T(v), a> \space = \space <<v,v>,a> \space$

That's better. I don't know if V is ℝ or ℂ in particular, we're supposed to assume it can be ℂ unless stated otherwise I believe.

So $<<v,v>,a> = \overline{a}<<v,v>,1> = \overline{a}<v,v> = <v, av> = <v, T^*(a)>$

$∴T^*(a) = av$

EDIT : If anyone sees this, I believe this is a Linear functional?

Last edited: Apr 7, 2013
18. Apr 8, 2013

### Fredrik

Staff Emeritus
<v,T*a>=<v,av> doesn't imply that T*a=av. Can you think of something you can do just a little bit differently that will allow you to conclude that T*a=av at the end of the calculation? (That is the correct formula for T*a).

A linear functional is a linear transformation into the field of scalars. Since the codomain of T* is V, not F, it's not a linear functional. It's just a linear transformation.

19. Apr 8, 2013

### Zondrina

This confused me, how does that not imply it? What could I have possibly done differently to arrive at the general formula for $T^*(a)$?

20. Apr 8, 2013

### Fredrik

Staff Emeritus
Take the vector space $\mathbb R^2$ for example. We have $$\langle (1,0),(2,0)\rangle=2 =\langle (1,0),(2,5)\rangle$$ but $(2,0)\neq (2,5)$. So it's not always true that if $\langle x,y\rangle=\langle x,z\rangle$, then $y=z$.

Also note that for all w such that <w,v>=0, we have <v,T*a+w>=<v,T*a>=<v,av>. This observation should be enough to cast some doubt on your derivation of T*a=av.

Last edited: Apr 8, 2013
21. Apr 8, 2013

### Zondrina

I understand what you're saying, what I still don't seem to understand is why after the manipulation I can't conclude that T*(a) = av? I understand there are counterexamples... does that mean I have to restrict something?

Last edited: Apr 8, 2013
22. Apr 9, 2013

### Fredrik

Staff Emeritus
Since <v,av> is equal to both <v,T*a> and <v,T*a+w>, why should the conclusion be T*a=av and not T*a=av+w with w≠0?

You need to go back to the beginning. $v$ is some specific member of $V$, which is used to define $T$. $T:V\to F$ is defined by $Tu=\langle u,v\rangle_V$ for all $u\in V$. This map is linear because you're using the convention that if V is an inner product space over ℂ, then the inner product $\langle,\rangle_V$ is linear in the first variable. (Physicists like to make it linear in the second variable instead).

It's true that this implies that for all $a\in F$, we have
$$\langle v,T^*a\rangle_V =\langle Tv,a\rangle_F =(Tv)a^* =\langle v,v\rangle_V a^* =\langle v,av\rangle_V.$$ But is this really all that the assumptions imply? Did you really have to insert the specific vector $v$ in the first slot of the inner product of the left? What if you insert something else there?

23. Apr 9, 2013

### Zondrina

Well... if i insert $u + w$ then :

$$\langle u+w,T^*(a) \rangle_V =\langle T(u+w),a\rangle_F =(T(u) + T(w))a^* =(\langle u,v\rangle + \langle w,v \rangle)_V a^* =(\langle u,av\rangle + \langle w,av\rangle)_V.$$

24. Apr 9, 2013

### Fredrik

Staff Emeritus
Right. The right-hand side is equal to $\langle u+w,av\rangle_V$, and you didn't make any assumptions about u or w. So it seems that no matter what vector you put in the first slot of the inner product that you start with, you still get a similar result. Is there any way you can use that to your advantage?