I believe the title contains all useful information for the problem. I tried to show that this map would have to be an isometry, but I can't seem to ensure that orthogonal vectors in the domain remain orthogonal in the image.
Asked
Active
Viewed 43 times
0
-
Such a linear map is norm-preserving and thus injective, and thus by rank-nullity, it is an isomorphism. By the polarization identity, norm-preserving and linear implies inner product-preserving. So, coupled with being an isomorphism, you have a unitary map. The key is the polarization identity (which in the real case is just a fancy way of saying $ab=\frac{(a+b)^2-(a-b)^2}{4}$, and in the complex case is essentially this with a few more $i$’s). – peek-a-boo Aug 31 '23 at 13:06
-
Thank you, I realized from your comment that it must be norm-preserving since any vector can just be scaled down to be a unit vector then scaled back after applying the linear map. – Daniel Mandragona Aug 31 '23 at 13:16
-
I would add that isometries preserve triangles so norm preserving implies it will also preserve orthogonality. – CyclotomicField Aug 31 '23 at 13:26
-
Does this answer your question? $T$ preserves inner product iff $|T\alpha |=|\alpha|$ or this? Confirmation of correctness in proof regarding norm preserving operator or that? relations between distance-preserving, norm-preserving, and inner product-preserving maps or other similar posts? – Anne Bauval Aug 31 '23 at 13:43