# Homework Help: Double transpose of a linear transformation

1. Feb 16, 2013

### SqueeSpleen

I'm using a book that has a loot of errors (luckly most of them are easy to recognize, like a = instead of a ≠ or viceversa, but some are way more serious), and I'm not sure if it's a new error or a thing I don't understand.
Either I didn't understood all the steps of the proof or the correct equality is:
$\Gamma (f(X)) = f^{tt}(X)$
$\Gamma (f(X)) = f^{tt}(\Gamma(X))$

http://img844.imageshack.us/img844/1603/bookkx.png [Broken]
Attemp to translation:
for all X $\in$ V (proof: being both members functions of W* in K, let $\psi \in$ W* then,
* I think they forgot to close the parenthesis.

Gamma is defined as the explicit isomorfism from V to V**
PD: There's a way of avoiding having to erase the post template every time you click the preview button?

Last edited by a moderator: May 6, 2017
2. Feb 16, 2013

### tiny-tim

Hi SqueeSpleen!
if it was ftt(X), then ftt would have to act on V, which it doesn't (and it would be represented by a diagonal arrow)

also, which line of the proof do you think should not have a Gamma?
if only life were that simple!

3. Feb 16, 2013

### SqueeSpleen

I thought that because:
Gamma goes from V to V** or W to W**
Transpose goes from:
Hom(V,W) to Hom(W*,V*)
If f goes from V to W, the tranpose goes from W* to V*
If we use it again, we got:
(V,W)tt=(W*,V*)t=(V**,W**)
And gamma of f goes from V** to W**

I understood this, so the equality is right, but I still don't get the whole proof; I'm not really sure why the gamma appears after the third equal sign, but I think it's the same reason why it dissapears after the first equal.
(I thought I understood it 2 days ago, I re-read it and I thought I didn't really previously understood it... I think I'd better sleep after messing everything up).

I deteled this fragment and I'll post it below so the thread I'll be a little easier to follow to others.

Last edited: Feb 16, 2013
4. Feb 16, 2013

### tiny-tim

you keep changing this!!

yes, that's correct … both the first and the third equals signs come from the definition of Gamma

5. Feb 16, 2013

### SqueeSpleen

$\Gamma_{v}:V^{*}\rightarrow K$
$\Gamma_{v}(\varphi )=\varphi (V)$
$\Gamma : V \rightarrow V^{**}$
$\Gamma(V) = \Gamma_{v}$
$f^{t}: W^{*} \rightarrow V^{*}, f^{t}(\varphi )=\varphi f$

$(\Gamma (f(X)))(\psi )=$
We apply the definition of gamma, letting f(X) be "v"
$=(\Gamma_{(f(X))})(\psi )=$
We apply the definition of gamma_v, letting f(X) be "v"
$=\psi (f(X))=$
We apply the definition of gamma_v, letting X be the "v"
$=\Gamma_{X}(\psi f)=$
We apply the definition of gamma, letting X be the "v"
$=\Gamma(X)(\psi f)=$
We apply the definition of transpose with ψ being the "$\varphi$"
$=\Gamma(X)(f^{t}(\psi))=$
Now, as $f^{t}$ goes from W* to V*, it's tranpose goes from V** to W**, so this time $\Gamma(X)$ is the
"$\varphi$" of the definition of tranpose.
Finally:
$=(f^{tt}(\Gamma(X)))(\psi )$

I think I earlier forget there were 2 differents gammas and messed up everything.

Last edited: Feb 16, 2013
6. Feb 16, 2013

### tiny-tim

yes … to put it simply:

just as ft(ψ) = ψf,

so ftt(Γ(X)) = (ft)t(Γ(X)) = Γ(X)(ft)

(and now insert a ψ, and you get the last-but-one equals sign)

7. Feb 16, 2013

### SqueeSpleen

I don't know if you saw my last edit.
It's everything ok? I'll handwrite it if it's everything ok :P

8. Feb 16, 2013

### tiny-tim

i'm not sure what you mean by "being the φ"
… when you apply the definition of transpose, you are interchanging (transposing!) two things

the first time, the two things were ψ and f … ψf = ftψ

the second time they were Γ(X) and ft … Γ(X)ft = (ft)tΓ(X)

9. Feb 16, 2013

### SqueeSpleen

$=\Gamma(X)(f^{t}(\psi))=$
We play a little with the parenthesis
$=(\Gamma(X)f^{t})(\psi)=$
Now, as $f^{t}$ goes from W* to V*, it's tranpose goes from V** to W**, so this time $\Gamma(X)$ is the
"$\varphi$" of the definition of tranpose.
Finally:
$=(f^{tt}(\Gamma(X)))(\psi )$

Now it's ok? If it isn't please explain me because I fail to see what's wrong.

10. Feb 17, 2013

### tiny-tim

(just got up :zzz:)

it's not wrong, but you can miss out the whole line …
and just write …
$=(\Gamma(X)f^{t})(\psi)$
$=(f^{tt}(\Gamma(X)))(\psi )$

(if you feel an explanation is needed, simply write "by definition of transpose")

11. Feb 17, 2013

### SqueeSpleen

Sorry :P
But I'm being a bit paranoid after my professor said one proof I made in a test, that used:
http://imageshack.us/photo/my-images/32/demostracionh.png/
I used this lemma and proved it, and she said it wasn't necessary because we can use the things saw on the book, so she didn't even bothered to check if it was right.
And I also used this:
http://imageshack.us/photo/my-images/202/observacion.png/
And the whole proof was wrong because I didn't explained why $V^{o}=0$
So after that, this is my face when I write things for linear algebra:

I try to specify as much as possible why I'm doing everything.

12. Feb 17, 2013

### tiny-tim

yes, but most of your extra line isn't an explanation of why the next line works, it's more a description of the consequences…
… the only reason why the line works is that ab = bta, and everything else is superfluous

in an exam, if you write too much, the examiner will think that you don't really understand what the reason is, and that you're just writing everything you know in the hope that it contains the important part!

in maths proofs, you get marks for being concise

13. Feb 17, 2013

### SqueeSpleen

I think that English+maths was too much, it took me 3 replies to understand what you were trying to say me xD
Thank you again, I'll try to be more concise here (I usually are in handwritting because I hate to handwrite, math was my favourite subject in elementary school because it was the one I needed to write the least...), while I love to write in the computer, and I'll stop right know because I don't want to spam anymore.

Last edited: Feb 17, 2013