Am I not understanding something, or does Axiom #2 not hold up when you have a vector of the form V=[a,ai] where i is the imaginary number and a is any real number
|V| = = a*.a + (ai)*.ai = a.a + (-ai).(ai) = a.a + a.a = 2a^2 Hence |V| = 0 iff a = 0 , and so |V| = 0 iff V = 0 QED (as long as the characteristic of the base field is not 2; wich is of course not the case with the real numbers field R as the base field)
You are a Professional. The best explanation one can get for the topic.
I think property 4, you get the complex conjugate of alpha so = α*
This is true if alpha is in multiplication with the bra vector v, if it is in multiplication with ket vector u then it comes out as it is
Amazing
Maybe it is unwise to use the * notation when dealing with vector spaces since that confuses the * notation with matrices.
Than You!🙏🙏😍
Excellent!
Is he a peruvian physicist?
He is a professor at MIT. en.wikipedia.org/wiki/Barton_Zwiebach
Thanks
thank you sir
Am I not understanding something, or does Axiom #2 not hold up when you have a vector of the form V=[a,ai] where i is the imaginary number and a is any real number
|V| = = a*.a + (ai)*.ai = a.a + (-ai).(ai) = a.a + a.a = 2a^2
Hence |V| = 0 iff a = 0 , and so |V| = 0 iff V = 0
QED (as long as the characteristic of the base field is not 2; wich is of course not the case with the real numbers field R as the base field)
Really??
ν=cu
??
nasa an America is so smart
sounds like she need more analog