Grassmann's Identity/Proof 1

Outline
Starting from a basis of $M \cap N$, we complete it to a basis of $M$ and one of $N$.

We then verify that the union of these basis is a basis of $M + N$.

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 is Less Than its Superspace this is finite-dimensional.

By Results concerning Generators and Bases of Vector Spaces, there exist non-empty 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 $\sequence {b_m}$ and $\sequence {d_p}$ be sequences of distinct vectors such that $B \cup C = \set {b_1, \ldots, b_m}$ and $D = \set {d_1, \ldots, d_p}$.

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 $k \in \closedint 1 p$.

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

and therefore $\lambda_j = 0$ for all $j \in \closedint 1 m$.

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

Thus we have:

Hence the result.