# V is a vector space such that $V = A\oplus A^\perp$ also $V = A \oplus C$ then can we say that $A^\perp = C$?

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!) – Mariano Suárez-Alvarez May 17 '12 at 7:42
@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? – srijan May 17 '12 at 7:51
@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. – Marc van Leeuwen May 17 '12 at 8:59
If, on the other hand, $A\oplus B=A\oplus C$ and $B\subseteq C$, then $B=C$. – rschwieb May 17 '12 at 11:19
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$$\left(Span\left\{\binom{1}{0}\right\}\right)^\perp=\left(Span\left\{\binom{0}{1}\right\}\right)^\perp\neq Span\left\{\binom{1}{1}\right\}$$
Rescue? ${}{}{}$ – talmid Jun 23 '12 at 0:09