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Question: let $W_1, W_2, W_3$ be three distinct subspaces of vector space $\mathbb{R^{12}}$ each having dimension $4$, and let $W=W_1 ∩W_2 ∩ W_3$ then which of the following is/ are correct?

(1) $W$ is subspace of $\mathbb{R^{12}}$

(2) $dim(W)< 8$

(3) $dim(W)>7$

(4) $dim(w)<10$

My attempt: clearly $W$ will be subspace of $V$ (as intersection of subspaces is again subspace) hence (1) is true.

I stuck on other options, but I know $dim(W)≤dimV=12$. Further, I know if $W_1, W_2$ are subspaces of finite dimensional vector space then,

$dim(W_1+W_2)= dim(W_1)+dim(W_2)-dim(W_1∩W_2)$

But, how to use this formula, as we had intersection of three subspaces? further, using above formula is useful here? Please help needed.

Akash Patalwanshi
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    the dimension of the intersection of subspaces is at most the smallest dimension of any of the subspaces involved. – Thomas Sep 28 '18 at 14:32
  • @Thomas sir, first of all thank you so much for your comment. Further is there is way of calculating dimension of the intersection of three subspaces. The above formula that I had mentioned can be used to calculate dimension of $W_1∩W2$ but how can we use it, to calculate dimension of intersection of three subspaces? – Akash Patalwanshi Sep 28 '18 at 14:38
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    https://math.stackexchange.com/questions/2065003/dimw-1-cap-w-2-cap-w-3-for-given-9-dimension-sub-spaces-w-1-w-2 Very closed to your question – Shweta Aggrawal Sep 28 '18 at 14:44

2 Answers2

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$W$ is also a subspace of $W_i$, so $\dim W \leq \dim W_i = 4$.

Rodrigo Dias
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  • sir, thank you so much for your answer, Further is there is way of calculating dimension of the intersection of three subspaces. The above formula that I had mentioned can be used to calculate dimension of $W_1∩W2$ but how can we use it, to calculate dimension of intersection of three subspaces? – Akash Patalwanshi Sep 28 '18 at 14:41
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    See here or here or even here for some related discussion – Rodrigo Dias Sep 28 '18 at 17:17
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$\dim W \leq \dim W_i = 4$ but since each $W_i$ are distinct $\dim W≠4$, therefore dimension of $W$ is strictly less than 4.Also notice that each subspace $W_1,W_2$ and $W_3$ can be chosen such that there intersection is $\{0\}$.Hence $$ 0≤\dim W \leq 3$$ Now obviously $W$ is subspace of $\mathbb{R}^{12}$, therefore correct options are $(1),(2)$ and $(4)$

Ibrahim Islam
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