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 $x\neq 0$, $Ax=0\,\Rightarrow A=0$.