Grassmann's Identity

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

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

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

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

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

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

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:

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