Adjoint of a linear operator (2)

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SUMMARY

The discussion focuses on proving two key properties of linear operators in finite-dimensional inner product spaces. Specifically, it establishes that the image of the adjoint operator T* is equal to the orthogonal complement of the kernel of T, denoted as im(T*) = (ker T)⊥. Additionally, it confirms that the rank of T is equal to the rank of its adjoint, rank(T) = rank(T*). These results are fundamental in understanding the relationships between linear transformations and their adjoints.

PREREQUISITES
  • Understanding of finite-dimensional inner product spaces
  • Knowledge of linear operators and their properties
  • Familiarity with the concepts of kernel and image of a linear transformation
  • Basic grasp of orthogonal complements in vector spaces
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  • Study the properties of adjoint operators in linear algebra
  • Learn about the Rank-Nullity Theorem and its applications
  • Explore orthogonal complements and their significance in inner product spaces
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Mathematicians, students of linear algebra, and anyone studying functional analysis will benefit from this discussion, particularly those interested in the properties of linear operators and their adjoints.

kingwinner
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Q: Suppose V is a finite dimensional inner product space and T:V->V a linear operator.
a) Prove im(T*)=(ker T)^(|)
b) Prove rank(T)=rank(T*)


Note: ^(|) is orthogonal complement

For this question, I don't even know how to start, so it would be nice if someone can give me some hints. Thank you!
 
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T* is the linear transformation such that, for all u and v, <Tu,v>= <u,T*v> . ker T= {v, Tv= 0). The "orthogonal complement" of ker T is {u |<u, v>= 0 for all v in ker T}. If w is in the image of T* then w= T*v for some v. Then <u, w>= <u,T*v>= <Tu, v> and if u is in the kernal of T, that is equal to what? That proves that im(T*) is a subset of the orthogonal complement of ker T. Prove the other way similarly. Of course, you should know that the rank(T)+ nullity(T)= dimension of V. (Nullity(T)= dimension of kernal of T and rank(T)= dimension of im(T).
 

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