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?


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

2) How can I complete the proof?

Thanks in advance!

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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: 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 $U^\perp$.

  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ that is not what i said. it is only $f_U(w)$ if $w \in W$. $\endgroup$ – supinf Jul 4 '17 at 13:20

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