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The question: Let Y be a finite-dimensional inner product space and T a linear operator on Y. Show that the range of T* is the orthogonal complement of the null space of T

Think i got one way:

took v$\in$Im(T*) and u$\in$Ker(T)

will show $(u|v)=0$

v$\in$Im(T*) => there is v'$\in$Y for it T*(v') = v

(u|T*(v')) = (T(u)|v') = (0|v') = 0

and this is true for any u$\in$Ker(T) => v is in the complement of Ker(T)

Julien
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dave
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1 Answers1

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What do you know about $\dim(\ker(T))$ and $\dim({\rm Im}(T)$? What do you know about $\dim({\rm Im}(T))$ and $\dim({\rm Im}(T^*))$?

ncmathsadist
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  • dim(Im(T)) + dim(ker(T)) = dimY,dim(KerT) + dim(complement of kerT) = dimY => dim(ImT) = dim(complement of KerT) – dave Mar 31 '13 at 01:48
  • Want to show that dim(Im(T*)) = dim(Im(T)) Cannot figure out the way.. – dave Mar 31 '13 at 01:52
  • ok i think i got it, since we know the matrix of T* is the conjugate transpose of the matrix of T we know r(T) = r(T) => dim(Im(T)) = dim(Im(T)). Thank you. – dave Mar 31 '13 at 05:31