Can You Solve These Linear Operator Problems in Inner Product Spaces?

[Defennder]

Homework Statement

1. Let T be a linear operator on an inner product space V. Let U = TT*. Prove that U = TU*.

2. For a linear operator T on an inner product space V, prove that T*T = T₀implies T = T₀.

Homework Equations

The Attempt at a Solution

1. This appeared at the outset to be a relatively simple one, but for some reason I can't prove it:

Let $$v,x \in V$$.
I know that U* = U, and
<v,Ux> = <v,TT*x>
= <T*v,T*x>

<v,TU*x> = <T*v,U*x>
=<T*v,TT*x>

So there's always an extra T I can't get rid of. How do I prove this?

2. I'll try to prove <v,T*Tx> = <v,T₀x>, knowing that (T*T)* = T*T
But as before this leads nowhere because there is again an extra T I can't get rid off:
<v,T*Tx> = <Tv,Tx> = <T₀v,x>

<v,Tx> (This one only has one T on the inner product).

 

[Dick]

It may appear simple at the outset, but you can't prove it, because it isn't even true. Let T=[[0,1],[1,0]]. T*=T. TT*=I=U. U*=U=I. U is not equal to TU*. Is there something about these linear operators you aren't telling us about?

 

[morphism]

Also, what is T_0, the zero operator?

 

[Defennder]

It may appear simple at the outset, but you can't prove it, because it isn't even true. Let T=[[0,1],[1,0]]. T*=T. TT*=I=U. U*=U=I. U is not equal to TU*. Is there something about these linear operators you aren't telling us about?

I'm not understanding your notation here. What does [[0,1],[1,0]] mean?

Also, what is T_0, the zero operator?

These are the problems exactly as they given:

6. Let T be a linear operator on an inner product space V. Let U₁ = T + T* and U₂ = TT*. Prove that U₁ = U₁* and U₂ = TU₂*.

It's easy to do the first part for U₁, but it's the U₂ linear operator I have trouble with.

[quote="Linear Algebra 3rd Edn pg 347 by Stephen Friedberg et al]11. For a linear operator T on anner product space V, prove that T*T = T₀ implies T = T₀. Is the same result true if we assume that TT* = T₀[/quote]Hey, T₀ actually refers to the zero transformation T₀: V -> W, T₀(x) = 0 for all $$x\in V$$. I didn't know that. Guess this is what happens when you skip chapters.

Now this can be done easily:
<v,T*Tx> = <Tv,Tx>= <v,T₀x> = <Tv,0>
Then by the "cancellation law": (<x,y> = <x,z> for all x implies y=z) (Question: does this cancellation also mean that <z,x> = <y,x> implies z=y? I think I can prove it but my proof may be faulty and I need to know if this is true.)
Tx=0 for all x which imples T=T₀

 

[Dick]

I'm not understanding your notation here. What does [[0,1],[1,0]] mean?

It means a 2x2 matrix whose first row is [0,1] and second row is [1,0]. U=TT*, U is not equal to TU*. For a general matrix det(U)=det(T)^2 and det(TU*)=det(T)^3. Are you sure that extra T is in the problem isn't a typo?

 

[Defennder]

Yeah I double-checked. The question is as stated.

 

[Dick]

Yeah I double-checked. The question is as stated.

Then did you figure out what's going on? Is there some assumption from the previous parts of the exercise you are supposed to carry into this one? Because it's certainly not true for general linear transformations.

 

[Defennder]

The question is standalone and the previous questions don't appear to have any relation to this one. I don't think it's a problem of notation either since T is used universally as a linear operator. Maybe it's a misprint or something.

 

[Dick]

Funny. It's an odd misprint to find in a 3rd edition.

 

[Defennder]

There's actually a 4th edn out there. But I haven't managed to find a second-hand copy of it yet, so I'm stuck with the 3rd edn on loan from the library since they only allow a limited 2 hour loan for their 4th edn.

 

[Dick]

Check the errata in the 4th edition. Look at the original question, i) U1=T+T*. Prove U1=U1*. Easy. ii) U2=TT*. The obvious question is 'prove U2=U2*'. The T came from nowhere.