Difficulty with EASY equivalence statement

[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 I am not meant to use the Polarization Identity or something like that. Can anyone help?

 

[George Jones]

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

 

[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 $S=T$

 

[George Jones]

Right, but surely I can't just say

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

Yes.

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

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

 

[George Jones]

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

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

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

and

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

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

 

[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.

 

[George Jones]

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

 

[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.

 

[George Jones]

See this http://www.math.gatech.edu/~green/Spring2002/Ma7334/spectral-theory.pdf, page 11 (pdf page 13).

Regards,
George

 

[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?

 

[Oxymoron]

I guess it is not trivial.

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

 

[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

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

 

[George Jones]

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

 

[Emieno]

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

 

[George Jones]

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

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

 

[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.

 

[George Jones]

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

 

[Emieno]

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

 

[Emieno]

The polarization identity says that

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

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

 

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

 

[Oxymoron]

Actually my version of the polarization identity is this:

Let $b\,:\,\mathcal{H} \times \mathcal{H} \rightarrow \mathbb{C}$ be a Hermitian bilinear form. Then define $\hat{b}\,:\, \mathcal{H} \rightarrow \mathbb{C}$ by $\hat{b}(x) = b(x,x)$. Then

$$b(x,y) = \frac{1}{4}\sum_{k=0}^3 i^k\hat{b}(x+i^k y) \quad \forall \, x,y \in \mathcal{H}$$

But I hate this version of the polarization identity - in fact I don't even understand it all that well.

 

[Oxymoron]

To save people from jumping back a page, here is a summary (no grAmmA+Ic@L errors).

1. Prove $$S=T \Rightarrow \langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle \quad \forall x \in \mathcal{H}$$

2. Prove $$\langle Sx\,|\,y \rangle = \langle Tx\,|\,y \rangle \Rightarrow S=T \quad \forall x,y \in \mathcal{H}$$

3. Prove $$\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle \Rightarrow \langle Sx\,|\,y \rangle = \langle Tx\,|\,y \rangle \quad \forall \, x,y \in \mathcal{H}$$.

Im pretty sure we did (1). We are doing (2) now, and hopefully (3) next.
The goal of this exercise is to prove that the following are equivalent.
1. $$S=T$$
2. $$\langle Sx\,|\,y \rangle \Rightarrow \langle Tx\,|\,y \rangle$$
3. $$\langle Sx\,|\,x \rangle \Rightarrow \langle Tx\,|\,x \rangle$$
for all $x,y \in \mathcal{H}$ and where $S,T \in \mathcal{B(H)}$.

 

[Oxymoron]

Well we came close in Post 20 but if x=0 then we can't assume A=0, and our hypothesis was for all x. So either a calculation error or we need to try something else.

 

[Emieno]

I find it a little bit weird as to why you go 123 not 132 ? The problem gets solved easier

 

[Emieno]

I mean in 3 we apply 1 and mention the reason as x component is arbitrary, from which we replace the right x with y to come to the second problem.

 

[Oxymoron]

Is it really that easy? Let me rearrange first...

1. $$S=T$$
2. $$\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle$$
3. $$\langle Sx\,|\,y \rangle = \langle Tx\,|\,y \rangle$$

for all etc, etc...

Now prove equivalence by 1 => 2 => 3 => 1?

 

[Emieno]

Finally in several previous posts, we mistook the point of proving such properties with the conceptions of finding other ordinary equations' roots, we can eliminate x,y without saying anything about the condition because we are trying to solve for or prove S=T not looking for roots of x and y. Something incorrect here anyway ?

 

[Emieno]

Ok, the third one can be solved as follows

<Sx|x>=<Tx|x> <-> <(S-T)x|x>=0

Now say as what I said in my post#25 to change back to <(S-T)x|y>=0 which also means <(Sx-Tx)|y> , then just go on again till you reach <Sx|y>=<Tx|y>. This is the second problem.
And then do as my post #27 mentioned...

 

[Oxymoron]

You are trying to prove (2) => (3) right?

 

[Oxymoron]

So we have to prove $\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle \Rightarrow \langle Sx\,|\,y \rangle = \langle Tx\,|\,y \rangle$.

$$(\Rightarrow)$$
Suppose $\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle$. Then rearrangining gives $\langle Sx\,|\,x \rangle - \langle Tx\,|\,x \rangle =0$ which implies $\langle (S-T)x\,|\,x \rangle = 0$. Now let $x=y$(?). Then $\langle (S-T)x\,|\,y \rangle = 0$. and reverse the process to solve.

Is this what you mean?