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How can I prove 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
up vote 4 down vote accepted

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

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 $\left<T^*v,w\right>=0$. And, indeed, $\left<T^*v,w\right>=\left<v,Tw\right>=\left<v,0\right>=0$.

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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

Let $A=ran(T^*), B=ker(T)^\bot$.

$A\subseteq B:$ For $x\in A, \ x=T^*y\ $ for some $y\in V$. Then, for any $z\in ker(T),$ $<x,z>=<T^*y,z>$

$=<y,Tz>=<y,0>=0.$ Hence $x\in B.$

$\\ \\ B\subseteq A:$ Because $V$ is finite dimensional and $A,B$ is subspace, it is equivalent to $A^\bot \subseteq B^\bot= ker(T)$. If $x\in A^\bot$, for any $y\in V$, $0=<x,T^*y>=<Tx,y>$ Therofore, $Tx=0$ (see exercise 8.1.1 (b)) , $x\in ker(T)$.

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Exercise 8.1.1 (b) is from what source? – Leucippus 23 hours ago

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