Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 2 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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