When can $0$ be a dimension of $(V \cap W)$?

Question

GRE 0568

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GRE 1268

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Why is that $0$ is included in the former but not in the latter?

It seems to me we just use the rule:

$$\dim(V+W)=\dim(V)+\dim(W)-\dim(V \cap W) \le \dim(\text{Well in this case we have X or} \ \mathbb R^4)$$

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What's up with Rambotutoring's solution then?

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Why doesn't the same argument apply to the latter?

Answer 1

Let me answer the question "Why doesn't the same argument apply to the latter?".

The argument that is being talked about is "a minimum of $0$ elements (corresponding to the case where $V$ and $W$ are disjoint except for the zero vector)".

That cannot happen in the case of two $4$-dimensional subspaces of a $7$-dimensional vectorspace: their intersection has at least dimension $1$. This follows from the dimension formula you quote yourself: $\dim(V) + \dim(W) - \dim(V \cap W) \leq \dim(X)$, so $8 - \dim(V \cap W) \leq 7$.

Answer 2

Well, there it is: in GRE 1268 we have

$$7=\dim X\ge\dim(V+W)=\dim V+\dim W-\dim(V\cap W)=8-\dim(V\cap W)\implies$$

$$\implies\dim(V\cap W)\ge8-7=1$$

but in GRE 0568 we have

$$4=\dim \Bbb R^4\ge\dim V+\dim W-\dim(V\cap W)=4-\dim(V\cap W)\implies$$

$$\implies \dim(V\cap W)\ge 0\;,\;\;\text{which is trivial, but also}$$

$$V\cap W\le V,\,W\implies \dim (V\cap W)\le\dim V=2$$

and from here the solution follows

Answer 3

Note that $ \dim(V) + \dim(W) -\dim(V+W) = \dim(V \cap W)$.

So we look at the cases individually: In the first case, $\dim(V)=2$, $\dim(W) = 2$ and $2 \leq \dim(V+W) \leq 4$. Therefore: $0 \leq \dim(V \cap W) \leq 2$.

In the second case, $\dim(V)=4$, $\dim(W) = 4$ and $4 \leq \dim(V+W) \leq 7$. Therefore: $1 \leq \dim(V \cap W) \leq 4$. So $\dim(V+W)$ cannot be zero.