# Proof TFAE

• Jul 5th 2006, 06:09 AM
TXGirl
Proof TFAE
Show that the following properties of an endomorphism d:V-->V of a vector space with a scalar product <.|.> are equivalent:

(i) d*=d^(-1)
(ii) <d(u)|d(v)>=<u|v> for all u,v elements of V
(iii) ||d(u)||= ||u|| for all u element of V
(iv) ||d(u) - d(v)|| = ||u-v|| for all u,v element of V

Note: If V is euclidean, a map d with the above properties is orthogonal. If V is unitary, then such a d is likewise unitary.
• Jul 5th 2006, 06:50 AM
ThePerfectHacker
Quote:

Originally Posted by TXGirl
Show that the following properties of an endomorphism d:V-->V of a vector space with a scalar product <.|.> are equivalent:

Note, sure I understand. How can you have a scalar product involving elements of a vector space!?!
• Jul 5th 2006, 07:05 AM
Rebesques
Quote:

with a scalar product <.|.> are equivalent:
I think she means inner product. Doesn't make sence otherwise :confused:

Anyway. Lets deal with the real case.

i)=>ii) <d(u)|d(v)>=<d*d(u)|v>=<d^{-1}d(u)|v>=<u|v>.

ii)=>iii) Just let u=v.

iii)=>iv) ||d(u)-d(v)||=||d(u-v)||=||u-v||.

iv)=>i) We will show that

a) <d*d(u)|v>=<u|v> for all u,v in V. This implies d*du=u for all u in V.

b) <dd*(u)|v>=<u|v> for all u,v in V. This implies dd*u=u for all u in V.

Now a) and b) imply d*=d^{-1}.

Lets prove a). We have ||d(u)-d(v)||^2=||u-v||^2 so

||d(u)||^2 - 2<du|dv> +||d(v)||^2 = ||u||^2 - 2<u|v> +||v||^2

and by using ||d(w)||=||w||,

<du|dv> = <u|v> or <d*d(u)|v>=<u|v>, and a) is proven.

Now b) is the same.
• Jul 5th 2006, 07:08 AM
TXGirl
well it just means that the scalar product operation is defined in the given space
what is the difference between saying the inner product and the scalar product??
• Jul 5th 2006, 07:22 AM
TXGirl
The assumptions here do not seem clearly founded to me:

i)=>ii) <d(u)|d(v)>=<d*d(u)|v>=<d^{-1}d(u)|v>=<u|v>.
__________________________________________________ ____

I said, we have that d* is the adjoint d.

So, <u|d(v)>=<d*(u)|v>

So d can then be understood as the identity function thus indicating that
<d(u)|d(v)>=<u|v> for all u,v elements of V.

Is that right??
• Jul 5th 2006, 10:59 AM
Rebesques
Quote:

what is the difference between saying the inner product and the scalar product??
The inner product is a generalization of the scalar product.

Quote:

So, <u|d(v)>=<d*(u)|v>
Use this identity with v=d(w) for some w to get ii from i.

Quote:

So d can then be understood as the identity function thus indicating that
<d(u)|d(v)>=<u|v> for all u,v elements of V.
Not true. This is actually the definition of a special operator called isometry. For example, in R^2, consider a rotation by a fixed angle. Then this mapping preserves the length of vectors, and so is an isometry.