1. ## 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.

2. 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!?!

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

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.

4. 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??

5. 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>.
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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??

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

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

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.