3
$\begingroup$

Let V be a finite dimensional vector space over the field $K$. Let $U,W$ be the subspaces, and assume that $V$ is the direct sum $U\oplus W$. Show that $V^*$ is equal to the direct sum $U^{\perp}\oplus W^{\perp}$.

I am aware the dual space has the same dimension. "Theorem:The map of $v\to L_v$ of $V$ into $V^*$ is an isomorphism."

I guess $U^{\perp}=W$ and $W^{\perp}=U$, so $U^{\perp}\oplus W^{\perp}=U^{\perp}+W^{\perp}$.

But unless $K=1$

I am not seeing how $U^{\perp}\oplus W^{\perp}\in V^{*}$ because that would imply V*=V, right?

Questions:

1) Is $V=V^{*}$?

2) How can I complete the proof?

Thanks in advance!

$\endgroup$
  • 1
    $\begingroup$ Referencing the orthogonal complement $U^{\perp}$ does not make sense unless $V$ is an inner product space. Is this meant to be the case? $\endgroup$ – Alex Mathers Jul 2 '17 at 18:42
  • 3
    $\begingroup$ If you look at the definition of $U^{\perp}$, it is almost certainly defined to be $$U^{\perp} = \{ f \in V^{\ast} : f(u) = 0 \textrm{ for all } u \in U \}$$ $\endgroup$ – D_S Jul 2 '17 at 18:43
  • $\begingroup$ @D_S ah, that makes much more sense. $\endgroup$ – Alex Mathers Jul 2 '17 at 18:46
  • 2
    $\begingroup$ In this sense, the notation $U^0$ is to be preferred. $\endgroup$ – Bernard Jul 2 '17 at 18:50
  • $\begingroup$ @D_S Are you saying that V must be zero, I mean all its elements zero? $\endgroup$ – Pedro Gomes Jul 3 '17 at 11:05
1
$\begingroup$

1: You are not wrong with the statement $V=V^*$, but they are only congorphic, so i think it would be better to write as $V \cong V^*$. For the purpose of the second question, this statement is not useful and i suggest you ignore it.

2: Here i will use the definition $$ U^\perp = \{f\in V^*: f(u)=0 \forall u\in U\}. $$ Using this definition (which is very common!) we can see that $U^\perp\subset V^*$.

For the direct sum you have to show two things: First, that $U^\perp \cap W^\perp=\{0\}$. Second, that $V^*=U^\perp+W^\perp$.

The first property is easier to show (try it). For the second property, let $f\in V^*$ be given. Note that $f$ is uniquely defined by its values on $U$ and $W$.

So we can define $f_U,f_W \in V^*$ as $$ f_U(u+w) = f(u) \quad,\quad f_W(u+w) = f(w). $$ where $u \in U$ and $w\in W$. (it might be necessary to show that these are well-defined).

Then $f_U \in W^\perp$ and $f_W\in U^\perp$ (why?). Finally $f = f_U+f_W \in W^\perp +U^\perp$, thus completing the proof of question 2.

There might be easier proofs, depending on what tools you have available, but the above is rather elemtary and you can get maybe a better feeling for dual spaces and $U^\perp$.

$\endgroup$
  • $\begingroup$ Thanks for your answer! What do you mean by $f = f_U+f_W \in W^perp +U^\perp$? $\endgroup$ – Pedro Gomes Jul 3 '17 at 12:27
  • 1
    $\begingroup$ that was a typo. I mean that $f=f_U + f_W$ and $f_U+f_W \in W^\perp + U^\perp$. $\endgroup$ – supinf Jul 3 '17 at 12:30
  • $\begingroup$ Should $f_U(u+w) = f(u) \quad,\quad f_W(u+w) = f(w)$ be instead f_U(u+w) = f(w) \quad,\quad f_W(u+w) = f(u) once $f_U(u)=0\:\:\text{and}\:\:f_W(w)=0$. I am asking this because if f_U is a functional that acts only upon the subspace U and f_W upon the subspace W, then $V^{*}$ would be zero and $f = f_U+f_W \notin W^\perp +U^\perp$? right? $\endgroup$ – Pedro Gomes Jul 3 '17 at 17:37
  • $\begingroup$ $f_U$ as defined in my answer acts on $V$, not just $U$. But $f_U$ is zero on $W$. $\endgroup$ – supinf Jul 4 '17 at 9:10
  • 1
    $\begingroup$ that is not what i said. it is only $f_U(w)$ if $w \in W$. $\endgroup$ – supinf Jul 4 '17 at 13:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.