Double dual/Double Transpose Question

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In summary, we have shown that if X" is identified with X and U" is identified with U, then T" = T. This is proven by showing that for any f in X", T"(f) maps L to Lu, which is the double dual of u. Therefore, T" = T, and the two linear homogeneous maps are equivalent.
  • #1
nyisles131
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The question that I am stuck on is:
Show that if X" (double dual of X) is identified with X and U" (double dual of U) with U via the duality relation, then T" (double transpose) = T.
(Duality relation is f(L) = L (x) where f is in X", L is in X', and x is in X)

So far, here is my work:

We know:

T: X --> U is a linear homogeneous map
Therefore,
T': U' --> X' where U' is the dual of U and X' is the dual of X
Then,
T": X" --> U" where X" is the double dual of X and U" is the double dual of U.

Also, X" is isomorphic to X, and U" is isomorphic to U.

I am missing something here, however. This is where I am stuck. How can one deduce that, in fact, T" = T? How do we show that two linear homogenoue maps are the equivalent?
The idea of a double dual has left me slightly confused and any help would REALLY be appreciated.

Thanks.
 
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  • #2
How do we show that two linear homogenoue maps are the equivalent?
You can show two linear maps are equal in exactly the same way as any other function: by proving that T(x) = T"(x) for all x.
 
  • #3
Let me check with you if this is correct...

Let T(x) = u
Then, we have to show that T"(f) = u where f is in X"?

We know the following:
For f in X", L in X', T"(f)(L) = fT'(L)
This implies: T"(f)(L) = T' (Lx) (by duality relation)
Then, T"(f)(L) = LT(x)
Then, T"(f)(L) = Lu
But, this does not equal u. What am I doing wrong here?
 
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  • #4
Let T(x) = u
Then, we have to show that T"(f) = u where f is in X"?
Strictly speaking, just as f is assumed to be the double dual of x, you need to show that T"(f) is the double dual of u.


We know the following:
For f in X", L in X', T"(f)(L) = fT'(L)
That's wrong.

T"(f) is an element of U".
L is an element of X'.
Therefore, one cannot evaluate T"(f) at L.
 
  • #5
I'm very confused right now.
By definition, isn't (T"(f))(L) = f(T'(l)?...Can you give me any suggestions as to how to approach this problem?
 
  • #6
Short answer: you're using something from the wrong space. You need to consider an element of U', since T"(f) is an element of U".



Honestly, I too find functions of functions confusing. (And worse, you're dealing with functions of functions of functions!)

My solution is simply to try and be extra precise and write everything down that I know -- in particular, I try to (at least mentally) write down what set everything lives in, and if it's a function, I write down what it's domain and its range is.

One cute little trick that works in this particular example is to have elements of X" act on the right. In particular, you would write:

(L)f

and not

f(L)

This has the benefit of the suggestively similar notation:

Lf = Lx

I'm not going to do that in what follows, though.


(Changing letters to reduce possible confusion)

If you have a map:

S : Y --> Z

then you have a map

S' : Z' --> Y'

Suppose g is an element of Z'. Then S'(g) is an element of Y'. In particular:

S'(g) : Y ---> F (where F is your base field)

So, S'(g) has to take something in Y as its argument. (Not something in Z -- that's the mistake you were making)

And, IIRC, the definition is:

S'(g)(y) = g(S(y))

(look at the type of everything, and make sure that the whole expression makes sense. e.g. note that S(y) is an element of Z)


(P.S. you have to assume that X and U are finite dimensional spaces. I just want to make sure you know that)
 
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  • #7
Ok, so let me start over again...
(FORGET ALL THE VARIABLES I USED BEFORE)

I have:
T: X --> U
T': U' --> X'
T": X" --> U"

Then, take f in X"
Therefore, T"(f) is in U".
Then, T"(f)(L) = f(T'(L)) for L in U'
Then T"(f)(L) = f(LT) because T'(L) = LT
Thus, T"(f)(L) = LT(x) by duality relation
Finally, T"(f)(L) = Lu, which is in U" by the duality relation
So, T"(f) does indeed map to Lu, which is in U".

Now, is it enough to say that since X" is isomorphic to X (dim X" = dim X) and U" is isomorphic to U (dim U" = dim U), then T" = T. Or is there a step I am missing here? (I feel like I am missing something, but I'm not sure)
 
  • #8
Is there anything else I need to do for this problem other than what I stated in the previous post?
 
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  • #9
Does anyone know if what I have stated 2 posts ago is the correct answer?
 
  • #10
Finally, T"(f)(L) = Lu, which is in U" by the duality relation
So, T"(f) does indeed map to Lu, which is in U".
This last line is correct. T"(f) isn't mapped to Lu. (T"(f) maps L to Lu) The important thing is that, since L is arbitrary, T"(f) is the dual of u.

So, you've proven that if T(x) = u, then T"(x") = u". When they say:

X" is identified with X​

they mean that (wave hands a bit) we are considering x" = x to be an actual equality. (similarly, that u" = u) And that is what's required to show T" = T.


If you feel uncomfortable with that level of imprecision, it should be okay to simply remember that:
T"(x") = u" iff T(x) = u
 
  • #11
I finally get it...:smile:
Thanks for your help!
 

Related to Double dual/Double Transpose Question

What is a double dual/double transpose question?

A double dual or double transpose question is a concept from mathematics and linear algebra. It refers to the process of taking the dual of a dual space, or the transpose of a transpose matrix. This results in the original space or matrix.

Why is the concept of double dual/double transpose important?

The concept of double dual/double transpose is important because it helps us understand the relationship between a space and its dual space, or a matrix and its transpose. It also has practical applications in fields such as physics, engineering, and computer science.

How is a double dual/double transpose question solved?

A double dual/double transpose question is solved by using the properties of duality and transpose. For example, to find the double dual of a vector space, we take the dual of its dual, which results in the original space. To find the double transpose of a matrix, we take the transpose of its transpose, which also results in the original matrix.

What is the difference between a double dual and a double transpose?

The main difference between a double dual and a double transpose is that they are defined in different contexts. A double dual is defined for vector spaces, while a double transpose is defined for matrices. However, they both follow the same concept of taking the "inverse" operation twice to result in the original object.

What are some real-world applications of double dual/double transpose?

Double dual/double transpose has many real-world applications. In physics, it is used to describe the relationship between different physical quantities and their duals. In engineering, it is used in circuit analysis and control systems. In computer science, it is used in data compression and signal processing. It also has applications in optimization and statistics.

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