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I have a vector space $V$ such that $V = A\oplus A^\perp$ i.e. $V$ is a direct sum of its subspace $A$ and orthogonal complement of $A$. Suppose we also have $V = A \oplus C$

Then can we say that $A^\perp = C$. If not then in what condition this relation may hold true? I think both subspace will have same dimension but i am not sure about equality of sub spaces.

I am confused here and need a clarification.

Thanks

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    Find yourself an example of all this with $V=\mathbb R^2$. (And do not look back here until you do... because someone is going to answer and ruin your fun!)2012-05-17
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    @MarianoSuárez-Alvarez I think result need not true. $\mathbb{R}^2$ can be written as a direct sum of X and Y axis also here Y axis can be replaced by any line passing through origin and still the direct sum will hold. Am i right?2012-05-17
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    Yes. ${}{}{}{}{}$2012-05-17
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    (Well, not **any** line: it cannot be the X axis!)2012-05-17
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    @MarianoSuárez-Alvarez Thank you very much. Suddenly i got confused with this but your hint helped me to clear my doubt.2012-05-17
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    @MarianoSuárez-Alvarez Oh yes.2012-05-17
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    What is the difference between $A \oplus B$ and $A + B$?2012-05-17
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    Please do not forget to write an answer answering your question! :)2012-05-17
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    @copper.hat $\oplus$ denotes outer sum and $+$ denotes inner sum, which only makes sense if you consider $A,B$ subsets of the same vector space. For all practical purposes they're the same thing, just different notations, since we can consider $A$ and $B$ subsets of $A\oplus B$.2012-05-17
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    @AlexBecker: Thanks.2012-05-17
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    @copper.hat: In this context (two subspaces of the same space) $A\oplus B$ denotes the same subspace as $A+B$, but _in addition_ expreses the fact that the sum is direct: $A\cap B=\{0\}$. If that condition is not satisfied one simply should not write $A\oplus B$ at all. Stated differently, one has $A\oplus B=A+B$ if and only if the left-hand-side is defined.2012-05-17
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    @AlexBecker: it is true that for unrelated spaces $A\oplus B$ is a freshly constructed space that contains _copies_ of $A$ and $B$ as subspaces, which subspaces in addition form a direct sum equal to the whole space. And in this setting it is $A+B$ that is not defined. However using this meaning of $\oplus$ when $A,B$ are subspaces of $V$ (as they are here) would be very confusing, especially if $A\cap B\neq\{0\}$. Such notation should not be used unless inevitable and clearly signalled. Note: nonzero $v\in A\cap B$ would give _two distinct_ elements of $A\oplus B$, one in $A$ another in $B$.2012-05-17
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    If, on the other hand, $A\oplus B=A\oplus C$ and $B\subseteq C$, then $B=C$.2012-05-17
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    @rschwieb That's what i wanted to confirm? Thanks2012-05-17
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    In the OP, if you additionally assume $A^\perp\subseteq C$ or $C\subseteq A^\perp$, then yes, equality holds. But as you have learned from previous comments, it's easy to find counterexamples when you don't have either of these.2012-05-17
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    **Please** remember to write your own answer to this question.2012-05-18
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    @MarianoSuárez-Alvarez Ok sir i will write.2012-05-18

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Let's rescate this question from the Unanswered Questions's limbo: $$\mathbb R^2\cong Span\left\{\binom{1}{0}\right\}\oplus Span\left\{\binom{0}{1}\right\}\cong Span\left\{\binom{1}{0}\right\}\oplus Span\left\{\binom{1}{1}\right\}$$and of course$$\left(Span\left\{\binom{1}{0}\right\}\right)^\perp=\left(Span\left\{\binom{0}{1}\right\}\right)^\perp\neq Span\left\{\binom{1}{1}\right\}$$

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    Rescue? ${}{}{}$2012-06-23
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    In english, yes. In spanish is "rescate" and in hebrew הצלה . I'd rather let the spanish version...:)2012-06-23
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    I know :-) I thought it was a typo.2012-06-23
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    @DonAntonio thank you very much.2012-06-23
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    Oh, it was a typo, @talmid...I was just chatting with my brother in Mexico (I'm in Israel) and the language's neurons of my brain (all 2 of them) got confused and went into strike.2012-06-23