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Let V be an inner product space, with T being a linear operator on V. How do I prove that $R(T^{*})^\perp =N(T)$? I tried setting $x\in R(T^{*})$ and $Ty\in N(T)$, and set up an inner product = 0 since $Ty\in N(T)$ but just got $T^{*}x\perp y$, and I don't know where to go from there.

Edit: and how could I prove it for a more general case $T:V\rightarrow W$?

George
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  • what do you mean by "more general case"? It seems Math1000 has given a proof... – Alex Mathers Mar 01 '16 at 00:23
  • For when T is a linear function from V->W, rather than just on V. I thought the inner product wouldn't necessarily apply, since it's possible for Tx/Ty to be in a different space, though I'm not sure. – George Mar 01 '16 at 00:27

1 Answers1

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For $T: V \rightarrow W$, we have: \begin{align} y\in N(T)&\iff \langle x,Ty\rangle_W=0, \ \forall x\in W\\ &\iff \langle T^*x,y\rangle_V = 0, \ \forall x\in W \\ &\iff y\in R(T^*)^\perp. \end{align}

Christopher A. Wong
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Math1000
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  • How would I prove it for a more general case T:V->W? – George Feb 29 '16 at 04:04
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    @George, I have modified the symbols to indicate how it works for the general case $T: V \rightarrow W$. In particular, the proof is not different, you just have to know which space, and what inner product (whether on $W$ or $V$), you're working with. – Christopher A. Wong Mar 01 '16 at 00:38
  • Thanks for the edit @ChristopherA.Wong – Math1000 Mar 01 '16 at 00:43