# Collection of Subspaces of a Vector Space

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi everyone, Here's a question I am struggling with recently. Hope you can give me some hints or ideas on how to solve this.

Question:

If the collection of subspaces of the $$K$$-vector space $$V$$ satisfies either distributive law $$A+(B\cap C)=(A+B)\cap (A+C)$$ or $$A\cap (B+C)=(A\cap B)+(A\cap C)$$, show that $$\mbox{dim}_{k}V\leq 1$$.

#### Sudharaka

##### Well-known member
MHB Math Helper
I think I have a solution. It would be nice if somebody could confirm its correctness. It is trivial that $$\mbox{dim}_{K}V=0$$ when $$V=\left\{0\right\}$$. Take any two elements $$v_1\mbox{ and }v_2\in V$$. Then consider the subspaces spanned by these elements. Let, $$A=<v_1+v_2>,\, B=<v_1>,\,C=<v_2>$$

Let us check whether these subspaces satisfy the first distributive law,

$A+(B\cap C)=A\Rightarrow \mbox{dim}(A+(B\cap C))=1$

and (note that all the pairwise intersections of these subspaces are trivial),

\begin{eqnarray}

\mbox{dim}\left[(A+B)\cap (A+C)\right]&=&\mbox{dim}(A+B)+\mbox{dim}(A+C)-\mbox{dim}(A+B+C)\\

&=&\mbox{dim}(A+B)+\mbox{dim}(A+C)-\mbox{dim}(B+C)\\

&=&\mbox{dim}(A)+\mbox{dim}(B)+\mbox{dim}(A)+\mbox{dim}(C)-\left(\mbox{dim}(B)+\mbox{dim}(C)\right)\\

\therefore \mbox{dim}\left[(A+B)\cap (A+C)\right]&=& 2

\end{eqnarray}

Hence these particular subspaces do not satisfy the first distributive law. Then they should satisfy the second distributive law.

$A\cap(B+C)=A$

and,

$(A\cap B)+(A\cap C)=\{0\}$

Therefore, $$A=\{0\}$$ and this implies, $$v_1=-v_2$$. Since we have chosen $$v_1\mbox{ and }v_2$$ to be arbitrary elements this means that $$V$$ is generated by one element. That is, $$\mbox{dim}_{K}V=1$$.

$\therefore \mbox{dim}_{K}V\leq 1$

#### Deveno

##### Well-known member
MHB Math Scholar
There are a couple of problems with your proof:

1. There is no guarantee $v_1,v_2$ are distinct, non-zero, or linearly independent. For example, if $V = F_2$, there is only one non-zero vector.

2. I fail to see how you can assume the intersection of your 3 spaces is pairwise trivial. This certainly is NOT true if $\text{dim}_{K}(V) = 1$ and $v_1,v_2$ are non-zero.

$\text{dim}[(A+B)\cap(A+C)] = 2$ seems severely flawed, and seems to rest on the assumption that $V$ is of dimension at least 2, with 2 linearly independent vectors $v_1,v_2$.

A similar problem exists with asserting that:

$(A \cap B) + (A \cap C) = \{0\}$.

If your conclusion (that $V$ is of dimension 1 or less) is indeed true, that will not be the case, unless both of $v_1,v_2$ are 0.

#### Sudharaka

##### Well-known member
MHB Math Helper
There are a couple of problems with your proof:

1. There is no guarantee $v_1,v_2$ are distinct, non-zero, or linearly independent. For example, if $V = F_2$, there is only one non-zero vector.

2. I fail to see how you can assume the intersection of your 3 spaces is pairwise trivial. This certainly is NOT true if $\text{dim}_{K}(V) = 1$ and $v_1,v_2$ are non-zero.

$\text{dim}[(A+B)\cap(A+C)] = 2$ seems severely flawed, and seems to rest on the assumption that $V$ is of dimension at least 2, with 2 linearly independent vectors $v_1,v_2$.

A similar problem exists with asserting that:

$(A \cap B) + (A \cap C) = \{0\}$.

If your conclusion (that $V$ is of dimension 1 or less) is indeed true, that will not be the case, unless both of $v_1,v_2$ are 0.
Thank you so much for pointing out the flaws in my proof. Let me explain how I thought about it.

Yes, I basically assumed that $$V$$ contains at least two distinct, non-zero, linearly independent elements. If $$V$$ contains only the element $$0$$ then it's dimension is obviously zero. If $$V$$ contains just two elements as you suggested (0 and another non-zero element), it's dimension would be one. These are trivial cases and we know that the inequality is satisfied in all these cases. So I just wrote the line "It is trivial that $$\mbox{dim}_{K}V=0$$ when $$V=\{0\}$$" to put these details under the carpet. #### Deveno

##### Well-known member
MHB Math Scholar
My point is, $V$ could be a field, in which case we cannot pick 2 linearly independent vectors at all. This is NOT the same as saying $V = \{0\}$.

In fact, we might have that $V$ is the smallest field possible, the galois field of order 2, in which case $V$ contains only ONE non-zero element (and $V \neq \{0\})$, which turns out to be its multiplicative identity.

What I think you need to do, is show that if $\text{dim}_K(V) \leq 1$, there is nothing to prove. Then, by way of forcing a contradiction, assume that one of the distributive laws holds, and show that if this holds, and $\text{dim}_K(V) \geq 2$, we have a contradiction to a certain (linearly independent) choice of $v_1,v_2$.

Pay CAREFUL ATTENTION to the case $\text{dim}_K(V) = 2$.

In this case, what you need to do is show that:

$\langle v_1,v_2\rangle = \langle v_1+v_2,v_1\rangle = \langle v_1+v_2,v_2\rangle$.

This is what will allow you to get specific values for the dimensions of the subspaces you want.

If $V$ were assumed finite-dimensional, you could without loss of generality, take $V$ to be:

$K^{\text{dim}_K(V)}$ and choose $v_1 = e_1$, $v_2 = e_2$.

But generally, using the axiom of choice, we could take some basis of $V$ (we actually NEED the axiom of choice to assume we HAVE a basis, this is something of a subtle point for infinite-dimensional spaces), and pick any two distinct basis elements. If $V$ is not finite-dimensional, it might be better to argue that $V$ must have at least 2 linearly independent elements (or else $V$ only has dimension 1, and we have nothing to prove, as indicated above). This avoids having to invoke the axiom of choice, which is rather like using a sledgehammer to swat a fly, for this particular problem.

The "heart" of your proof DOES work, but you left a "hole" for the 1-dimensional case, and proofs shouldn't HAVE holes.