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show TB is linear if B^2=0 Tb(A)=(A+B)^2-(A+2B)(A-2B)
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Multiply out the brackets, and you should come out wiser :D
TB(A)=A^2 +2BA+B^2-[A^2-4B^2] =2BA
when B^2=0
how does it show its linear though?
Anything non-linear would have higher powers of A or B.
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Or indeed negative powers of -2 or lesser.
ok, thanks, so if B=\[\lceil 0 1 \rceil \lfloor 0 0 \rfloor\] (dont know how to do matrices) but with 0 1 on the top and 00 on the bottom) what is the rank of B?
\[\lceil 0 1 \rceil\] \[\lfloor 0 0 \rfloor\]
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