# Dimension of the sum of subspaces

Let $V_1, \ldots, V_n$ be $n$ subspaces of a vector space $V$.

Is there a formula for $\dim(V_1 + \cdots + V_n)$ similar to $\dim(V_1 + V_2)=\dim(V_1) + \dim(V_1) - \dim(V_1 \cap V_2)$?

-
– lhf Jan 26 '12 at 12:30
Didn't notice those, thanks for referring. – Peter P. Jan 26 '12 at 13:24
As a curiosity: At the moment the answer related to this is the most upvoted in the MO thread Examples of common false beliefs in mathematics – Martin Sleziak Jan 26 '12 at 16:09

This example will illustrate the difficulty.

Let $i,j,k$ be the standard basis vectors for 3-dimensional real space. Let $W,X,Y,Z$ be the subspaces spanned by $i,j,k,i+j$, respectively. Then $W+X+Y$ has dimension 3, $W+X+Z$ has dimension 2, but $W,X,Y,Z$ all have dimension 1, and any intersection of two or more has dimension 0. So, no formula using dimensions of the four spaces and their intersections can distinguish between $W+X+Y$ and $W+X+Z$.

-

You can do it iteratively using $$\dim(V_1 + V_2) = \dim(V_1) + \dim(V_2) - \dim(V_1 \cap V_2)$$ i.e., set $V_1' = V_1$, then $V_2' = V_1' + V_2$, so you get $$\dim(V_2') = \dim(V_1') + \dim(V_2) - \dim(V_1' \cap V_2)$$ and so on, let $V_i' = V_{i-1}' + V_i$ so $$\dim(V_i') = \dim(V_{i-1}') + \dim(V_i) - \dim(V_{i-1}' \cap V_i)$$

If you want to, you can substitute the formula for $\dim (V_{i-1}')$ in the method above to get a formula, e.g. if you have $V_1, V_2, V_3$: $$\dim(V_2') = \dim(V_1) + \dim(V_2) - \dim(V_1 \cap V_2)$$ \begin{align*} \dim(V_3') &= \dim(V_2') + \dim(V_3) - \dim(V_2' \cap V_3) = \\ &=\dim(V_1) + \dim(V_2) - \dim(V_1 \cap V_2) + \dim(V_3) - \dim( (V_1 + V_2) \cap V_3) \end{align*} and of course $V_3' = V_1 + V_2 + V_3$, so this yields a formula for the dimension. The same method can be applied to an arbitrary finite number of subspaces, which yields \begin{align*} \dim \left( \sum_{i = 1}^n V_i \right) &= \dim \left( \sum_{i = 1}^{n-1} V_i \right) + \dim(V_n) - \dim \left( \sum_{i = 1}^{n-1} V_i \cap V_n \right) = \\ &= \dots = \sum_{i = 1}^n \dim(V_i) - \sum_{j = 1}^{n-1} \dim \left( \sum_{k = 1}^j V_k \cap V_{j+1} \right) \end{align*}

-
In the last line, surely you mean finite number of subspaces? – Ravi Jan 26 '12 at 15:22
Yeah, sure. Corrected. – Calle Jan 26 '12 at 15:25
In the last line, why do we have $- \dim \left(\sum_{i=1}^{n-1} V_i \cap V_n \right)$ instead of $- \dim \left( V_n\cap\sum_{i=1}^{n-1} V_i\right)$? – Leon Feb 8 at 22:55
@Leon, they are the same thing. – Calle Feb 9 at 22:35