theorem on adjoint of linear transformation

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November 5th 2011, 09:46 AM
abhishekkgp

theorem on adjoint of linear transformation

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

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

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

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

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

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

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

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

Can someone show the reverse containment??
November 5th 2011, 11:35 AM
FernandoRevilla

Re: theorem on adjoint of linear transformation

Quote:

Originally Posted by abhishekkgp

Can someone show the reverse containment??

Hint Use that $V,\;W$ are finite dimensional vector spaces.
November 5th 2011, 08:24 PM
abhishekkgp

Re: theorem on adjoint of linear transformation

Quote:

Originally Posted by FernandoRevilla

Hint Use that $V,\;W$ are finite dimensional vector spaces.

Let $\{e_1,e_2,\ldots,e_m\}$ be an orthogonal basis of $\text{null }} T$ .
Let $\{e_1,e_2, \ldots,e_n \}$ be its extension to an orthogonal basis of $V$ .
So $\{e_{m+1},e_{m+2},\ldots,e_n\}$ is an orthogonal basis of $(\text{null } T)^{\perp}$ .

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

Can you please again drop a hint?
I know i have to use the fact that W is also finite dimensional.
November 5th 2011, 09:00 PM
Deveno

Re: theorem on adjoint of linear transformation

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

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

then $\left<Ty,w\right> = 0$ , 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),

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

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

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

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

now take the perp of both sides.
November 5th 2011, 09:41 PM
abhishekkgp

Re: theorem on adjoint of linear transformation

Quote:

Originally Posted by Deveno

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

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

then $\left<Ty,w\right> = 0$ , 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),

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

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

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

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

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.