# Linear algebra proof involving subspaces and dimensions.

Let $W_1$ and $W_2$ be subspaces of a finite-dimensional vector space $V$. Determine necessary and sufficient conditions on $W_1$ and $W_2$ so that $\dim(W_1 ∩ W_2) = \dim(W_1)$.

Sorry if my post looked like a demand. My English is poor so I copied the exercise out of the book and did not change the wording of the question because I did not want to change the meaning (as I did in my last post). I was only requesting help/guidance and was not making any demands. I am studying the subject myself so its not a "homework" problem. Thank you to everyone who posted and helped.

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The dimension counting formula may come in handy:

$\operatorname{dim} (W_1 + W_2) = \operatorname{dim} (W_1) + \operatorname{dim} (W_2) - \operatorname{dim} (W_1 \cap W_2).$

Hence if $\operatorname{dim} (W_1) = \operatorname{dim} (W_1 \cap W_2)$, we have that $\operatorname{dim}(W_1 + W_2) = \operatorname{dim} W_2$ by the formula above. Now $W_1 + W_2$ is defined to be the subspace of $V$ whose elements are of the form $w_1 + w_2$ where $w_i \in W_i$. Now how can this formula hold? If $W_1$ and $W_2$ intersect trivially the sum of the two spaces becomes direct so that in fact we are calculating $\operatorname{dim}(W_1 \oplus W_2) = \operatorname{dim} W_1 + \operatorname{dim} W_2$ .

Plugging this into the formula $\operatorname{dim} (W_1 + W_2) = \operatorname{dim} W_2$ we get that $\operatorname{dim} W_2 = 0$. So now if we don't want degenerate cases like that we may assume that $W_1$ and $W_2$ have non-trivial intersection. Your job now is to tell me why $\operatorname{dim} (W_1 + W_2) = \operatorname{dim} W_2$ implies that $W_1 \subset W_2$ assuming $W_2$ is not the zero space.

Hint: If $U$ is a subspace of $V$ and $\operatorname{dim} U = \operatorname{dim} V,$ then $U = V$.

Therefore you have found a necessary condition for your formula to hold. Is this condition sufficient? Well it should be clear from the dimension counting formula above because then $\operatorname{dim} (W_1 + W_2) = \operatorname{dim} W_2$ so that cancelling terms on both sides of the formula and rearranging we get

$$\operatorname{dim} W_1 = \operatorname{dim} (W_1 \cap W_2).$$

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I am assuming your question is not a homework problem. – user38268 Feb 8 '12 at 5:14
Benjamin, that seems a mighty big assumption – user16299 Feb 8 '12 at 7:12
@YemonChoi That it is not a homework problem? – user38268 Feb 8 '12 at 7:19
Yes. Why do you assume it is not? Everything here points towards homework, I would say. – Did Feb 8 '12 at 8:25
@DidierPiau Well usually I like to give the OP the benefit of the doubt. – user38268 Feb 8 '12 at 9:18

HINT: Suppose that $\dim(W_1\cap W_2)=m$. Take a basis $B$ for $W_1\cap W_2$; it has $m$ members. These members must be linearly independent (why?). Now if $\dim(W_1)$ is also $m$, so that $B$ is a linearly independent set of $m$ vectors in $W_1$, is $B$ a basis for $W_1$?

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$\bf Hint:$ $\operatorname{dim}(W_1\cap W_2)=\operatorname{dim}(W_1)$ if and only if $W_1\subset W_2$

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The dimension of the intersection is less or equal than the dimension of each subspace since it is a subspace of each one. If the dimension of the intersection is equal to the W1 , that means : the intersection is equal to W1 and this means that W1 is contained or equal to W2. Then the condition you want is this (necessary and sufficient).

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