Question

show that T*T=0 implies T=0

Answer 1

@yummydum T is a matrix and T* is its conjugate transpose.

Answer 2

T*T is not equal to T times T. it's T*=conjugate transpose

Answer 3

lol oh okay...sorry...ill just back away slowly

Answer 4

@vbfn \[
\ \ \ (A^*)_{ij}=\overline{A_{ji}}\\
\text{Since }A_{ij}\text{ is complex, it can be written of the form }a+bi\\
\text{where }a,b\text{ are real. Let us define }a_{ij},b_{ij}\text{ such that}\\
\ \ \ A_{ij}=a_{Aij}+b_{Aij}i\\
\text{Therefore we reduce our original conjugate }\overline{A_{ji}}\text{ as follows.}\\
\ \ \ (A^*)_{ij}=a_{Aji}-b_{Aji}i\\
\text{Now, let's remind ourselves what the matrix product }AB\text{ really means.}\\
\text{If }A\text{ is an }m\text{-by-}n\text{ matrix and }B\text{ an }n\text{-by-}p\text{ matrix,}\\
AB\text{ is an }m\text{-by-}p\text{ matrix such that}\\
\ \ \ [AB]_{ij}=\sum_{r=1}^nA_{ir}B_{rj}\\
\text{Now, suppose }T\text{ is an }m\text{-by-}n\text{ matrix, therefore }T^*\text{ is an }n\text{-by-}m\\\text{matrix; }T^*T\text{ is then an }m\text{-by-}m\text{ matrix such that}\\
\ \ \ [T^*T]_{ij}=\sum_{r=1}^n(T^*)_{ir}T_{rj}\\
\ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{r=1}^n\overline{T_{ri}}T_{rj}\\
\ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{r=1}^n(a_{Tri}-b_{Tri}i)(a_{Trj}+b_{Trj}i)\\
\ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{r=1}^n[a_{Tri}a_{Trj}-a_{Trj}b_{Tri}i+a_{Tri}b_{Trj}i+b_{Tri}b_{Trj}]\\
\text{Given }T^*T=0\text{ it follows}\\
\ \ \ [T^*T]_{ij}=0\\
\ \ \ \sum_{r=1}^n[a_{Tri}a_{Trj}-a_{Trj}b_{Tri}i+a_{Tri}b_{Trj}i+b_{Tri}b_{Trj}]=0\\
\text{It is trivial to see this is true for }T=0\text{ and therefore}\\
\ \ \ a_{Tri}=a_{Trj}=b_{Tri}=b_{Trj}=0.\\
\text{Demonstrate this is true iff }T=0\text{; imagine that }\\
\ \ \ a_{Tri}a_{Trj}-a_{Trj}b_{Tri}i+a_{Tri}b_{Trj}i+b_{Tri}b_{Trj}\ne0\\
\text{for some }i,j,r.
\]