1. ## theorem on adjoint of linear transformation

Suppose $\displaystyle T:V \rightarrow W$ is a linear transformation, where V and W are finite dimensional inner product spaces. Prove that $\displaystyle \text{range } T^{*}=(\text{null } T)^{\perp}$.

I can prove it one direction:
Let $\displaystyle v \in \text{range } T^{*}$.

Then $\displaystyle v=T^{*}w$ for some $\displaystyle w \in W$.

Let $\displaystyle v_1 \in \text{null } T$.

Then $\displaystyle \left<v_1,T^{*}w \right>=\left<Tv_1,w \right>=\left<0,w \right>=0$.

So $\displaystyle \left< v_1, v \right>=0 \, \forall \, v_1 \in \text{null } T$.

Hence $\displaystyle v \in (\text{null } T)^{\perp}$.

We have $\displaystyle \text{range } T^{*} \subseteq (\text{null } T)^{\perp}$.

Can someone show the reverse containment??

2. ## Re: theorem on adjoint of linear transformation

Originally Posted by abhishekkgp
Can someone show the reverse containment??
Hint Use that $\displaystyle V,\;W$ are finite dimensional vector spaces.

3. ## Re: theorem on adjoint of linear transformation

Originally Posted by FernandoRevilla
Hint Use that $\displaystyle V,\;W$ are finite dimensional vector spaces.
Let $\displaystyle \{e_1,e_2,\ldots,e_m\}$ be an orthogonal basis of $\displaystyle \text{null }} T$.
Let $\displaystyle \{e_1,e_2, \ldots,e_n \}$ be its extension to an orthogonal basis of $\displaystyle V$.
So $\displaystyle \{e_{m+1},e_{m+2},\ldots,e_n\}$ is an orthogonal basis of $\displaystyle (\text{null } T)^{\perp}$.

I need to show that $\displaystyle \exists \, w \in W \text{ such that } T^{*} w= e_i, \, \text{ where } (m+1)\leq i \leq n$.

Can you please again drop a hint?
I know i have to use the fact that W is also finite dimensional.

4. ## Re: theorem on adjoint of linear transformation

suppose $\displaystyle y \in (\text{image }T^*)^\bot$

so $\displaystyle \left<y,T^*w\right> = 0$, for all w in W.

then $\displaystyle \left<Ty,w\right> = 0$, for all w in W.

so Ty = 0, so y is in null(T).

(note: why can we conclude that Ty = 0? well, we can always pick w = Ty, since Ty

is an element of W, and since <Ty,w> = 0 for all w in W, in particular,

<Ty,Ty> = 0, so Ty = 0).

on the other hand, if y is in null(T),

$\displaystyle \left<Ty,w\right> = \left<0,w\right> = 0$ for any w in W,

but $\displaystyle \left<Ty,w\right> =\left<y,T^*w\right>$

so we conclude that $\displaystyle y \in (\text{image }T^*)^\bot$

so $\displaystyle (\text{image }T^*)^\bot = \text{null }T$.

now take the perp of both sides.

5. ## Re: theorem on adjoint of linear transformation

Originally Posted by Deveno
suppose $\displaystyle y \in (\text{image }T^*)^\bot$

so $\displaystyle \left<y,T^*w\right> = 0$, for all w in W.

then $\displaystyle \left<Ty,w\right> = 0$, for all w in W.

so Ty = 0, so y is in null(T).

(note: why can we conclude that Ty = 0? well, we can always pick w = Ty, since Ty

is an element of W, and since <Ty,w> = 0 for all w in W, in particular,

<Ty,Ty> = 0, so Ty = 0).

on the other hand, if y is in null(T),

$\displaystyle \left<Ty,w\right> = \left<0,w\right> = 0$ for any w in W,

but $\displaystyle \left<Ty,w\right> =\left<y,T^*w\right>$

so we conclude that $\displaystyle y \in (\text{image }T^*)^\bot$

so $\displaystyle (\text{image }T^*)^\bot = \text{null }T$.

now take the perp of both sides.
thank you Denevo. that helped.
But i am still looking for a direct way to solve the problem.