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Adjoint operator

by typhoonss821
Tags: adjoint, operator
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typhoonss821
#1
Feb4-10, 03:42 AM
P: 14
I recently teach myself linear algebra with Friedberg's textbook.
And I have a question about adjoint operator, which is on p.367.

Definition Let T : V → W be a linear transformation where V and W are finite-dimensional inner product spaces with inner products <‧,‧> and <‧,‧>' respectively. A funtion T* : W → V is called an adjoint of T if <T(x),y>' = <x,T*(x)> for all x in V and y in W.

Then ,my question is how to prove that there is a unique adjoint T* of T ?

Can anyone give me some tips ? thanks^^
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radou
#2
Feb4-10, 04:02 AM
HW Helper
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P: 3,224
Assume that there is another adjoint transformation, let's say T**.
Mandark
#3
Feb4-10, 05:07 AM
P: 41
To show that there exists such a function, let [tex]v_1, \ldots , v_n[/tex] be an orthonormal basis for V, so that [tex]x = \sum_i \langle x,v_i\rangle v_i[/tex] for any x in V then we have for all x in V and y in W:

[tex]\langle T(x), y\rangle ' = \langle T (\sum_i \langle x,v_i\rangle v_i ), y\rangle '[/tex]
[tex] = \sum_i \langle x, v_i\rangle \langle T(v_i), y\rangle '[/tex]
[tex] = \langle x, \sum_i \overline{ \langle T(v_i),y\rangle '} v_i\rangle [/tex]
which is in the form that we'd like.

Which shows that [tex]T^*(y) = \sum_i \overline{\langle T(v_i),y\rangle '} v_i[/tex] for all y in W works.

typhoonss821
#4
Feb4-10, 06:04 AM
P: 14
Adjoint operator

really appreciate^^


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