Suppose $U$ and $V$ are $2$ dimensional subspaces of $\mathbb{R}^4$. How do you determine the dimension of $U \cap V$? I know that

\begin{equation*} \text{dim}(U + V) = \text{dim}(U)+ \text{dim}(V) - \text{dim}(U \cap V). \end{equation*}

So it seems that $\text{dim}(U \cap V) = 0$.

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    $\begingroup$ There is not enough information to uniquely determine the answer. The dimension of the intersection can be 0, 1, or 2. $\endgroup$ – Qiaochu Yuan Oct 8 '10 at 23:49
  • $\begingroup$ how can it be 1 or 2? $\endgroup$ – Tom Oct 8 '10 at 23:56
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    $\begingroup$ @Tom for 2, let U=V. For 1, consider U and V as planes which are intersecting in a line. $\endgroup$ – BBischof Oct 9 '10 at 2:03

First: it is incorrect, in general, to even write $\dim(U\cup V)$, because $U\cup V$ is almost never a subspace: it is a subspace if and only if $U\subseteq V$ or $V\subseteq U$. Not being a subspace, it doesn't even make sense to talk about its dimension.

Rather, what you probably meant is the correct equation $$\dim(U+V) = \dim(U) + \dim(V) - \dim(U\cap V)$$ (though I prefer to put the intersection on the left hand side, because in that form it is valid even in the infinite dimensional case). Here, $U+V$ is the smallest subspace that contains $U$ and $V$, and it happens to equal the set of all vectors of the form $u+v$ with $u\in U$ and $v\in V$.

The equation does not give you the full answer, because there are several situations that can occur: you could have $U$ and $V$ intersect trivially: this is what happens when $U+V=\mathbb{R}^4$. For an explicit example, you could have $U=\{(a,b,0,0):a,b\in\mathbb{R}\}$, and $V=\{(0,0,c,d) : c,d\in\mathbb{R}\}$. Here, $\dim(U\cap V)=0$.

Or you could have that $U$ and $V$ intersect in a one-dimensional subspace (for example, take $U$ as above, but take $V=\{(0,b,c,0) : b,c\in\mathbb{R}\}$).

Or you could have $U=V$, in which case the intersection has dimension $2$.

What you can say is: if the spaces are distinct, then the intersection will either have dimension 1 or dimension 0. But that is all you can say with the given information.

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If you have B and B', basis of U and V respectively, you can check if the vectors of B are linearly independent with those of B'. If so, then $\dim(U\cap V) = 0$. If not, you may be able to write a new basis of $U+V$ eliminating the linearly dependent vectors. $\dim(U\cap V)$ is the amount of vectors you have eliminated from $U+V$'s basis.

Also, $\dim(U+V) = \dim(U) + \dim(V) - \dim(U\cap V)$.

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  • $\begingroup$ Also, keep in mind that if $U \oplus V= \mathbb{R}^4$, then $dim(U \cap V)=0$ $\endgroup$ – F M Oct 8 '10 at 23:51
  • $\begingroup$ @Tom: It can be 2 if $U=V$. It can be 1 if one vector from U's basis is linearly dependent with the vectors from V's basis or vice versa. $\endgroup$ – F M Oct 9 '10 at 0:19
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    $\begingroup$ I edited just to make the "dim"s into roman (using \dim); pet peeve of mine inherited from my advisor (mathematical operators written with roman characters that are two characters or longer are written in roman, not italic; dim, ker, id, rk, ord, etc). $\endgroup$ – Arturo Magidin Oct 9 '10 at 2:38

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