Annihilator of a Direct Sum: Proving V0=U0⊕W0 for V=U⊕W

[Adgorn]

Homework Statement

Suppose V=U⊕W. Prove that V⁰=U⁰⊕W⁰. (V⁰= annihilator of V).

Homework Equations

(U+W)⁰=U⁰∩W⁰

The Attempt at a Solution

Well, I don't see how this is possible. If V⁰=U⁰⊕W⁰, then U⁰∩W⁰={0}, and since (U+W)⁰=U⁰∩W⁰, it means (U+W)⁰={0}, but V=U⊕W, so V⁰={0}. I don't think this is the desirable result so it means I misunderstood something along the way, clarification would be appreciated.

 

[andrewkirk]

(U+W)⁰=U⁰∩W⁰

This part is not relevant. It is not the case that $V=U+W$. We have $V=U\oplus W$ which has a different meaning.

To get a better handle on this, think about the following: say $u\in U^0-\{0_U\}$. What is a corresponding element of $V$ that is in $V^0$? Then do the same for $w\in W^0-\{0_W\}$.

 

[zwierz]

Suppose V=U⊕W. Prove that V0=U0⊕W0. (V0= annihilator of V).

that is wrong.
the correct conclusion must be $V^*=U^0\oplus W^0$
Indeed,
1) it is clear that $U^0\cap W^0=\{0\}$; it also follows from

(U+W)0=U0∩W0

2) take any function $f\in V^*$ and define a linear function $f_U$ is follows $f_U(x):=f(x)$ provided $x\in U$ and $f_U(x)=0$ provided $x\in W$;
similarly $f_W(x):=f(x)$ provided $x\in W$ and $f_W(x)=0$ provided $x\in U$.
Obviously it follows that $f=f_U+f_W,\quad f_U\in W^0,\quad f_W\in U^0$

 

[Adgorn]

This part is not relevant. It is not the case that $V=U+W$. We have $V=U\oplus W$ which has a different meaning.

To get a better handle on this, think about the following: say $u\in U^0-\{0_U\}$. What is a corresponding element of $V$ that is in $V^0$? Then do the same for $w\in W^0-\{0_W\}$.

But V=U⊕W means V=U+W and U∩W={0}, meaning the theorem still applies, doesn't it?
Also I am not familiar with the notation 0_u and 0_w, so i'd appreciate an explanation about that.

 

[Adgorn]

that is wrong.
the correct conclusion must be $V^*=U^0\oplus W^0$
Indeed,
1) it is clear that $U^0\cap W^0=\{0\}$; it also follows from

2) take any function $f\in V^*$ and define a linear function $f_U$ is follows $f_U(x):=f(x)$ provided $x\in U$ and $f_U(x)=0$ provided $x\in W$;
similarly $f_W(x):=f(x)$ provided $x\in W$ and $f_W(x)=0$ provided $x\in U$.
Obviously it follows that $f=f_U+f_W,\quad f_U\in W^0,\quad f_W\in U^0$

Well this does make sense, if it is indeed a misprint in the book it would not be the first one...