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How can I proof that, if $V$ is a finite-dimensional vector space with inner product and $T$ a linear operator in $V$, then the range of $T^*$ is the orthogonal complement of the null space of T?

I know what I must do (for a $v$ in the range of $T^*$, i have to show that $v\perp w$ for every $w$ in $Ker(T)$ and then do the opposite, but I don't know how to show that this inner product is zero!

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Is $T^*$ the traspose operator of $T$? –  Loronegro Mar 1 '13 at 21:36
    
@Loronegro it's the adjoint. –  user62182 Mar 1 '13 at 21:54

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In order to show that the range of $T^*$ is the orthogonal complement of $\ker T$, we have to show that $\forall v \in ImT^*$, $\forall w\in \ker T$: $<v,w>=0$.

Note that vectors in the range of $T^*$ are of the form $T^*v$ for $v\in V$. Now, let $w\in\ker T $. We have to show that $<T^*v,w>=0$. And, indeed, $<T^*v,w>=<v,Tw>=<v,0>=0$.

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You're using the fact that the range of $T^*$ is invariant under $T^*$ right? That is, if $\left\{ v_1,...,v_n\right\}$ is a basis for $V$ then $\left\{ T^* v_1,...,T^* v_n \right\}=\left\{ T^* v_1,...,T^* v_m\right\}$, with $m<n$, is a basis for the range, and $\left\{ T^* v_{m+1},...,T^* v_n\right\}$ a basis for the null space, right? –  user62182 Mar 1 '13 at 19:43
    
I'm not sure I understand the "invariant" part of what you said... But, anyway, I am not working with bases at all. I mean, I showed this part of your original statement: for every vector $v$ in the range of $T^*$, and every $w\in \ker T$, $v$ is orthogonal to $w$. –  Ludolila Mar 1 '13 at 19:50
    
I meant invariant under T. –  user62182 Mar 1 '13 at 20:07
    
Denote the range of $T^*$ by $ImT^*$. To say that $ImT^* $ is invariant under $T$, is to say that $\forall v \in ImT^*$ we have $Tv \in ImT^*$. And this is not something I want to claim/use. –  Ludolila Mar 1 '13 at 20:12
    
Regarding what you said about the basis: note that if $\{v_1,...,v_n\}$ is a basis for $V$, it doesn't mean that $\{T^*v_1,...,T^*v_m \}$ form a basis for $ImT^*$ (although these vectors do span the range, they are not necessarily linearly independent). –  Ludolila Mar 1 '13 at 20:15

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