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Math Help - Adjoint of linear operators

  1. #1
    Member Last_Singularity's Avatar
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    Adjoint of linear operators

    Question: Let T be a linear operator on a finite-dimensional vector space V. Show that:
    1. N(T^*T)=N(T)
    2. rank(T) = rank(T^*)
    3. rank(A^*A)=rank(AA^*)=rank(A)

    I figured that if I can prove (1), then it follows via the dimension theorem that rank(T^*T)=rank(TT^*)=rank(T). Then if I just find a basis, I can use the previous fact to solve (3). I just cannot get started on (1), though. And what about (2)? Any tips would be helpful - thanks!
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Last_Singularity View Post
    Question: Let T be a linear operator on a finite-dimensional vector space V. Show that:
    1. N(T^*T)=N(T)
    2. rank(T) = rank(T^*)
    3. rank(A^*A)=rank(AA^*)=rank(A)

    I figured that if I can prove (1), then it follows via the dimension theorem that rank(T^*T)=rank(TT^*)=rank(T). Then if I just find a basis, I can use the previous fact to solve (3). I just cannot get started on (1), though. And what about (2)? Any tips would be helpful - thanks!
    The space V has to be more than a vector space. It must be an inner-product space in order for the adjoint operation to be defined.

    For (1), it's clear that if Tx=0 then T*Tx=0. So N(T) ⊆ N(T*T). For the reverse inclusion, notice that T^*Tx=0\ \Longrightarrow\ 0 = \langle T^*Tx,x\rangle = \langle Tx,Tx\rangle = \|Tx\|^2 \Longrightarrow\  Tx=0.

    For (2), suppose T has (finite-dimensional) range W_0, and let W_1 be the orthogonal complement of W_0 in V. If x\in W_1 and y\in V then \langle T^*x,y\rangle = \langle x,Ty\rangle = 0, so that T^*(V) = T^*(W_0). Therefore \text{rank}(T^*) \leqslant \text{dim}(W_0) = \text{rank}(T). The same argument with T and T* interchanged gives the reverse inequality.
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