# Difficulty with EASY equivalence statement

1. Sep 29, 2005

### Oxymoron

I want to be able to prove that

$$S=T \, \Leftrightarrow \, \langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle$$

given that $S,T \in \mathcal{B(H)}$ and $x \in \mathcal{H}$.

If $S$ and $T$ are bounded linear operators and I suppose that $S = T$ then I have to be able to prove that

$$\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle$$

Now since $S = T$ then $\|S\| = \|T\|$ and in particular

$$\|Sx\| = \|Tx\| \quad \forall \, x \in \mathcal{H}$$.

Im not really sure how to progress from here. I hope Im not meant to use the Polarization Identity or something like that. Can anyone help?

2. Sep 29, 2005

### George Jones

Staff Emeritus
Maybe I'm missing something deep, but doesn't this follow trivially from the definition of equality of operators?

If $S$ and $T$ are bounded linear operators, then

$$S=T \, \Leftrightarrow \, Sx = Tx$$

for every $x$ in $\mathcal{H}$.

Regards,
George

3. Sep 29, 2005

### Oxymoron

Right, but surely I can't just say

Since $S=T \Leftrightarrow Sx=Tx$ then $\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle$.

This seems too easy. And how would I prove the other way, ie given that $\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle$ prove $S=T$

Last edited: Sep 29, 2005
4. Sep 29, 2005

### George Jones

Staff Emeritus
Yes.

Oops, I didn't see the arrow in the other direction - I'll have to think about this. Clearly, $\langle Sx\,|\,y \rangle = \langle Tx\,|\,y \rangle$ for every $x$ and $y$ iff $S=T$, but I don't know if this helps.

Regards,
George

5. Sep 29, 2005

### George Jones

Staff Emeritus
This doesn't appear to be true, i.e., I think I've found a counterexample.

Take $\mathcal{H}=\mathbb{R}^2$ with the standard inner product. Let

$$S = \left( \begin{array}{cc} 0 & 1\\ 0 & 0 \end{array} \right)$$

and

$$T = \left( \begin{array}{cc} 0 & 0\\ 1 & 0 \end{array} \right).$$

Then, $Sx = \left(x_2 , 0)^t$, $Tx = \left(0 , x_1)^t$, and $\langle Sx , x \rangle = \langle Tx , x \rangle = x_{1} x_{2}$ for every $x$.

Regards,
George

6. Sep 29, 2005

### Oxymoron

Well...as long as you have a Hilbert space (whose ground field is complex) then if two bounded linear operators $S,T \in \mathcal{B(H)}$ then $S = T$ is equivalent to saying that $\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle$

This is the exact wording of my problem.

7. Sep 29, 2005

### George Jones

Staff Emeritus
It seems that the fact the field is $\mathbb{C}$ is important. It looks like my counterexample doesn't work with $\mathbb{R}^2$ replaced by $\mathbb{C}^2$.

Regards,
George

8. Sep 29, 2005

### Oxymoron

Indeed George, the field must be complex.

Anyway, this was all leading to something. I want to actually prove that

$$\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle \quad \Rightarrow \quad \langle Sx\,|\,y \rangle = \langle Tx\,|\,y \rangle$$

for all $x,y \, \in \mathcal{H}$ and for all $S,T \, \in \mathcal{B(H)}$ where $\mathcal{H}$ is a Hilbert space over $\mathbb{C}$.

I am afraid that this is much more difficult and I have an idea that it will involve using the Polarization identity. But I need to sleep first.

Last edited: Sep 29, 2005
9. Sep 29, 2005

### George Jones

Staff Emeritus
See this link, page 11 (pdf page 13).

Regards,
George

10. Sep 30, 2005

### Oxymoron

Thanks for the link George, I like the introduction in that pdf.

I want to go on what you said about "clearly $\langle Sx\,|\,y \rangle = \langle Tx\,|\,y \rangle$ implies $S=T$".

I mean, this seems soooo obvious. But I want to justify it. Is the proof of this implication trivial or not?

11. Sep 30, 2005

### Oxymoron

I guess it is not trivial.

George, your pdf claims that proving this implication requires an EASY application of the Polarization identity.

12. Sep 30, 2005

### Oxymoron

I had a go at it, but as usual got nowhere.

I want to prove that

$$\langle Sx\,|\,y \rangle = \langle Tx\,|\,x \rangle \quad \Rightarrow \quad S=T$$

for all $x,y \in \mathcal{H}$ and for all $S,T \in \mathcal{B(H)}$.

So I began by supposing that $\langle Sx\,|\,y \rangle = \langle Tx\,|\,x \rangle$. I want to prove $S=T$ but proving $S-T=0$ will suffice.

Now we know that the orthogonal complement of a Hilbert space is 0 (and $S-T=0$), so it we must show that

$$S=0 \quad \Leftrightarrow \langle Sx\,|\,y \rangle = 0$$

for every $x,y \in \mathcal{H}$.

The polarization identity says that

$$4\langle Sx\,|\y \rangle = \langle S(x+y)\,|\,(x+y) \rangle - \langle S(x+y)\,|\,(x-y) \rangle + i\langle S(x+iy)\,|\,(x+iy) \rangle - i\langle S(x-iy)\,|\,(x-iy) \rangle$$.

13. Sep 30, 2005

### George Jones

Staff Emeritus
Let $A = S - T$. Then

$$\langle Sx\,|\,y \rangle = \langle Tx\,|\,y \rangle \Leftrightarrow \langle Ax\,|\,y \rangle = 0.$$

Fix $x$. Since the above is true for every $y$, $Ax=0$. But $x$ was arbitary, and thus $Ax=0$ for every $x$. Consequently, $A=0$.

Regards,
George

14. Sep 30, 2005

### Emieno

Do you need the condition whether x is not 0 since you say it is arbitrary?

15. Sep 30, 2005

### George Jones

Staff Emeritus
I meant to say it was arbitrary; actually, I said it was "arbitary."

I don't think any restriction needs to be placed on x.

Regards,
George

16. Sep 30, 2005

### Emieno

But the condition in the given problem is All x belongs to H, and what if x=0 ? A doesn't need to equal to o but still satisfies the condition Ax=0, I just mean that point.

17. Sep 30, 2005

### George Jones

Staff Emeritus
It's true that if Ax = 0 for every non-zero x, then A = 0, but the case x = 0 doesn't have to be excluded. If Ax = 0 for every x, then A = 0.

Regards,
George

18. Sep 30, 2005

### Emieno

If so, the original problem mistook about the condition, not All x,y in H satisfy the problem.

19. Sep 30, 2005

### Emieno

Is your formula correct ? Can you rewrite it if it were wrong since I don't have the book to check it now ?

20. Sep 30, 2005

### Oxymoron

Ok, so I suppose no Polarization identity needed then.

So we know that $\langle Sx\,|\,y \rangle = \langle Tx\,|\,y \rangle$

then rearrangining gives

$$\langle Sx\,|\,y \rangle - \langle Tx\,|\,y \rangle = 0$$

which implies

$$\langle (Sx-Tx)\,|\,y \rangle = 0$$

which implies

$$\langle (S-T)x\,|\,y \rangle = 0$$

Let $A = S-T$, then

$$\langle Ax\,|\,y \rangle = 0$$

Now fix $x \in \mathcal{H}$. Since the above is true for every $y \in \mathcal{H}$, then $Ax = 0$ for every $x$. Now if $x=0$ then this does not imply $Ax = 0$. So for every $x\neq 0$, $Ax=0\,\Rightarrow A=0$.