Dimension of sum of $3$ subspaces of a vector space.

Question

I would like to know if my proof is correct or not. I have seen other questions about this, but they do not cover what I am specifically asking. If $V$ is a vector space and $A$, $B$, and $C$ are subspaces of $V$, then $\dim\left(A+B+C\right)=\dim\left(A\right)+\dim\left(B\right)+\dim\left(C\right)$ implies that $V=0$.

I have $3$ cases: $\dim\left(\left(A+B\right)+C\right)$, $\dim\left(A+\left(B+C\right)\right)$, and $\dim\left(\left(A+C\right)+B\right)$.

I am using the formula $\dim\left(W_{1}+W_{2}\right)=\dim\left(W_{1}\right)+\dim\left(W_{2}\right)-\dim\left(W_{1}\cap W_{2}\right)$, applying it to $3$ subspaces, and equating it to the formula above.

For the first case, I get $$\dim\left(\left(A+B\right)+C\right)=\dim\left(A+B\right)+\dim\left(C\right)-\dim\left(\left(A+B\right)\cap C\right)$$ $$=\dim\left(A\right)+\dim\left(B\right)+\dim\left(C\right)-\dim\left(A\cap B\right)-\dim\left(\left(A+B\right)\cap C\right),$$ giving that $A\cap B=\left\{0\right\}$ and $\left(A+B\right)\cap C=\left\{0\right\}$. $A\cap B=\left\{0\right\}$ implies that $A+B$ is a direct sum, so $\left(A+B\right)\cap C=V\cap C=\left\{0\right\}$, so $C=\left\{0\right\}$.

For the second case, I get that $B\cap C=\left\{0\right\}$ and $A\cap\left(B+C\right)$. $B+C$ is a direct sum and $C=\left\{0\right\}$, so $B=V$, and we have $A\cap\left(B+C\right)=A\cap V=\left\{0\right\}$, so $A=\left\{0\right\}$.

For the third case, I get $A\cap C=\left\{0\right\}$, which means $A+C=V$, but $A+C=\left\{0\right\}+\left\{0\right\}$, so $V=\left\{0\right\}$.

Is this correct?

Answer 1

What you are trying to prove is not true. Take $V=\mathbb R^3$, and $$ A=\{(x,0,0):\ x\in\mathbb R\},\ \ B=\{(0,y,0):\ y\in \mathbb R\},\ \ C=\{(0,0,z):\ z\in\mathbb R\}. $$ Then $$ \dim(A+B+C)=3=1+1+1=\dim A+\dim B+\dim C. $$

The mistake you are repeatedly making is saying that $A\cap B=\{0\}$ implies $V=A+B$. If that were the case, every vector space would be two-dimensional.

Answer 2

Hmm, it looks like there are a few problems here, if I'm understanding your question correctly. The counterexample I'll use throughout here is $V = \mathbb{R}^3$, with $e_1 = (1,0,0)$, $e_2 = (0,1,0)$, and $e_3 = (0,0,1)$.

1. Why does $A \cap B = \{0\}$ imply that $A + B = V$? If $A = \text{span}\{e_1\}$ and $B = \text{span}\{e_2\}$ then $A \cap B = \{0\}$ but their direct sum is just a two-dimensional subspace of $V$.

2. Point (2) goes for your second and third cases as well -- $B\cap C = \{0\}$ doesn't imply $B + C = V$ and $A\cap C = \{0\}$ doesn't imply $A + C = V$.

EDIT: I reread your question (and comment) and removed one of the points.