0
$\begingroup$

I'm having trouble understanding the proof for existence of isometry between null(T) (same as null(|T|) and (range T)⊥. They do have the same dimension, but how can you guarantee the existence of a linear map that maps orthonormal vectors e1 to f1 (with no dependency on f2, f3, ...), e2 to f2 (with no dependency on f1, f3, ...), e3 to f3 (with no dependency on f1, f2, ...), ...

Shouldn't e1 be expressed as a linear combination of f1, f2, f3, ...

e1 = <e1.f1>f1 + <e1.f2>f2 + .....

excerpt from Linear Algebra Done Right

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.