# Dimension of Sum and Intersection of Vector Spaces

## Theorem

Let $\left({G, +_G, \circ}\right)_K$ be a $K$-vector space.

Let $M$ and $N$ be finite-dimensional subspaces of $G$.

Then $M + N$ and $M \cap N$ are finite-dimensional, and:

- $\dim \left({M + N}\right) + \dim \left({M \cap N}\right) = \dim \left({M}\right) + \dim \left({N}\right)$

## Proof

First, suppose $M \subseteq N$ or $N \subseteq M$.

Then the assertion is clear.

Assume that $M \cap N$ is a proper subspace of both $M$ and $N$.

Let $B$ be a basis of $M \cap N$.

By Dimension of Proper Subspace Less Than its Superspace this is finite-dimensional.

By Results concerning Generators and Bases of Vector Spaces, there exist nonempty sets $C$ and $D$ disjoint from $B$ such that:

- $B \cup C$ is a basis of $M$
- $B \cup D$ is a basis of $N$.

The space generated by $B \cup C \cup D$ contains both $M$ and $N$.

Hence it contains $M + N$.

But as $B \cup C \cup D \subseteq M \cup N$, the space it generates is contained in $M + N$.

Therefore $B \cup C \cup D$ is a generator for $M + N$.

If $d$ is a linear combination of $D$ and also of $B \cup C$, then $d \in M \cap N$.

So $d$ is a linear combination of $B$, and consequently $d = 0$ as $B \cup D$ is linearly independent and $D$ is disjoint from $B$.

In particular, $D$ is disjoint from $B \cup C$.

Next we show that $B \cup C \cup D$ is linearly independent and hence a basis of $M + N$.

Let $\left \langle {b_m} \right \rangle$ and $\left \langle {d_p} \right \rangle$ be sequences of distinct vectors such that $B \cup C = \left\{{b_1, \ldots, b_m}\right\}$ and $D = \left\{{d_1, \ldots, d_p}\right\}$.

Let $\displaystyle \sum_{j \mathop = 1}^m \lambda_j b_j + \sum_{k \mathop = 1}^p \mu_k d_k = 0$.

Then $\displaystyle \sum_{k \mathop = 1}^p \mu_k d_k = - \sum_{j \mathop = 1}^m \lambda_j b_j$.

Hence $\displaystyle \sum_{k \mathop = 1}^p \mu_k d_k$ is a linear combination of $D$ and also of $B \cup C$.

By the preceding, then $\displaystyle \sum_{k \mathop = 1}^p \mu_k d_k = 0$.

Hence $\mu_k = 0$ for all $k \in \left[{1 \,.\,.\, p}\right]$.

Thus: $\displaystyle \sum_{j \mathop = 1}^m \lambda_j b_j = 0$

and therefore $\lambda_j = 0$ for all $j \in \left[{1 \,.\,.\, m}\right]$.

Therefore $B \cup C \cup D$ is linearly independent.

Thus we have:

\(\displaystyle \dim \left({M + N}\right)\) | \(=\) | \(\displaystyle \left\vert{B \cup C \cup D}\right\vert\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left\vert{B \cup C}\right\vert + \left\vert{D}\right\vert\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left\vert{B \cup C}\right\vert + \left\vert{B \cup D}\right\vert - \left\vert{B}\right\vert\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dim \left({M}\right) + \dim \left({N}\right) - \dim \left({M \cap N}\right)\) |

$\blacksquare$

## Sources

- Seth Warner:
*Modern Algebra*(1965)... (previous)... (next): $\S 27$: Theorem $27.15$