# Direct sum of subspaces

I am rather confused. Suppose $V$ is a finite dimensional vector space and $A,B,C$ are (non-trivial) subspaces of $V$ such that $V=A\oplus B=A\oplus C=B\oplus C$ and it is said that there is a subspace of dimension 2 of $V$ whose intersection with any one of $A,B,C$ is a one dimensional subspace.

My confusion: Now since $V$ is a direct sum of these pairs of these subspaces this means that the pairwise intersection of $A,B,C$ has to be $\{0\}$. But then $A\oplus B=A\oplus C$ means that $B,C$ must be the same space, which is clearly wrong. What is going on here?

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Think about the case where $V$ is 2-dimensional and $A$, $B$ and $C$ are 1-dimensional (i.e., lines). It is not necessarily true that $B=C$. Draw some pictures. – Tom Cooney May 20 '12 at 22:08
@TomCooney: Ah, thank you! I am guessing that I can form the 2D space by picking one vector from $A$ an one from $B$. Is this correct? – Alagna May 20 '12 at 22:13

You can't 'subtract' the spaces, ie, $A\oplus B=A\oplus C$ does not mean $B=C$.

Concrete example: Choose $V=\mathbb{R}^2$. Take $A = \mathbb{sp}\{e_1\}$, $B = \mathbb{sp}\{e_2\}$, $C = \mathbb{sp}\{e_1+e_2\}$. Take $V$ as the two dimensional subspace in question.

More generally:

Choose $a,b$ to be non-zero elements of $A,B$ respectively. Let $S = \mathbb{sp}\{a,b\}$. Then it should be clear that $S$ is a 2-dimensional subspace of $V$, and that $A \cap S = \mathbb{sp}\{a\}$, $B \cap S = \mathbb{sp}\{b\}$. It remains to be shown that $S \cap C$ is 1-dimensional.

First, $S \cap C$ cannot be 2-dimensional, since if it was, we would have $a \in C$, which would contradict $V=A\oplus C$.

Finally, since $A\oplus B=A\oplus C$, we can write $a+b = \lambda a + \mu c$, where $\mu \neq 0$. Then we have $c = \frac{1}{\mu}((1-\lambda)a+b)$, and clearly $c \in S \cap C$. Hence $S \cap C$ is 1-dimensional.

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 Thank you, that is a good example! so if $A,B,C$ are unknown subspaces of some unspecified vector field $V$, would the sought-after 2D space be just the span of $\{a, b\}$ where $a\in A, b\in B$? – Alagna May 20 '12 at 22:23 @ArturoMagidin: Thank you, how might I find the space? Is there a way to visualize what is going on? – Alagna May 20 '12 at 22:38 @Alagna: Sorry; I might be wrong. I need to think about it a bit more; that choice definitely gives you the correct intersection with $A$ and with $B$. The intersection with $C$ is at most of dimension $1$; I thought it might be trivial, but on reflection it perhaps can't. – Arturo Magidin May 20 '12 at 22:41 @Alagna: As long as $a,b$ are non-zero elements of $A,B$ then the span of $a,b$ is the requisite 2-dimensional space. I have modified my answer a little, originally I thought you were looking for an example. – copper.hat May 20 '12 at 23:12 Thank you again, copper.hat! – Alagna May 20 '12 at 23:24