Note, sure I understand. How can you have a scalar product involving elements of a vector space!?!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:
(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.
I think she means inner product. Doesn't make sence otherwisewith a scalar product <.|.> are equivalent:
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.
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??
The inner product is a generalization of the scalar product.what is the difference between saying the inner product and the scalar product??
Use this identity with v=d(w) for some w to get ii from i.So, <u|d(v)>=<d*(u)|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.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.