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Math Help - inner product space, complex case

  1. #1
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    inner product space, thank you!

    i'd be extremely grateful if anyone can help me out with this
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    Suppose V is a finite dimensional inner product space and T : V → V a linear operator.

    1. Prove im(T∗ ) = (kerT )⊥.
    2. Prove rank(T ) = rank(T∗ ).


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    T∗ should be displayed as T^*, as is (kerT )⊥ should be (kerT)^⊥
    Last edited by ericmik; December 14th 2008 at 07:25 AM.
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  2. #2
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    Quote Originally Posted by ericmik View Post
    Suppose V is a finite dimensional inner product space and T : V → V a linear operator.

    1. Prove im(T∗ ) = (kerT )⊥.
    2. Prove rank(T ) = rank(T∗ ).
    From the relation \langle Tx,y\rangle = \langle x, T^*y\rangle it follows that if Tx=0 then x\perp T^*y. In other words, \text{ker}(T)\subseteq(\text{Im}(T^*))^\perp.

    The same argument with T and T* interchanged shows that \text{ker}(T^*)\subseteq(\text{Im}(T))^\perp. Therefore \text{null}(T^*)\leqslant n-\text{rank}(T) = \text{null}(T) by the rank plus nullity theorem, where n = dim(V). Interchanging T and T* again, you see that \text{null}(T)\leqslant  \text{null}(T^*), and therefore \text{null}(T) = \text{null}(T^*). That shows (by the rank plus nullity theorem again) that \text{rank}(T) = \text{rank}(T^*).

    Finally, it follows that \text{null}(T) = \text{rank}((T^*)^\perp). But we know that \text{ker}(T)\subseteq(\text{Im}(T^*))^\perp, and if these spaces have the same dimension then they must be equal.
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  3. #3
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    thank u so much
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