# theorem on adjoint of linear transformation

• Nov 5th 2011, 10:46 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 $\left=\left=\left<0,w \right>=0$.

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??
• Nov 5th 2011, 12:35 PM
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.
• Nov 5th 2011, 09:24 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.
• Nov 5th 2011, 10:00 PM
Deveno
Re: theorem on adjoint of linear transformation
suppose $y \in (\text{image }T^*)^\bot$

so $\left = 0$, for all w in W.

then $\left = 0$, 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),

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

but $\left =\left$

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.
• Nov 5th 2011, 10:41 PM
abhishekkgp
Re: theorem on adjoint of linear transformation
Quote:

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

so $\left = 0$, for all w in W.

then $\left = 0$, 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),

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

but $\left =\left$

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.