Question

show that T*T=0 implies T=0.
note: T* is the conjugate transpose of matrix T

Answer 1

if T = 0, obviously T*=0. but i don't know how to prove it.

Answer 2

\[\ \ \ z\bar{z}=0\\\ \ \ (a+bi)(a-bi)=0\\\ \ \ a^2+b^2=0\\\text{What possible values of}a \text{ and }b\text{ could result in that being zero? It'd require...}\\\ \ \ a^2=-b^2\\\text{Since }a\text{ and }b\text{ are real, the only possible values that would work here are zero.}\\\ \ \ 0^2=-0^2\\\text{Now, I expect you recognize that the conjugate transpose of a real matrix is very}\\\text{similar, since complex numbers can be treated as }2\times2\text{ real matrix. You can prove}\\\text{by extension pretty easily.}\]

Answer 3

I think I need to prove T*=T

Answer 4

I think once I can prove that T is a self-adjoint operator on V, then u and v are orthogonal; that is <u,v>=0
then I can prove T*T=0
no? yes?