Given a finite-dimensional affine space $(X,\vec X,\Theta)$, for any two affine subspaces $M$ and $N$, prove that $$\overrightarrow{\langle M\cup N\rangle}=\overrightarrow{M}+\overrightarrow{N}+K\vec{ab},$$ where $a$ is an arbitrary point in $M$, $b$ an arbitrary point in $N$, and $K$ the field of scalars. If $Y$ is any subset of $X$, $\langle Y\rangle$ is the notation for spanned subspace (smallest subspace containing $Y$). If $Y$ is an affine subspace of $X$, $\overrightarrow{Y}$ denotes the direction of the affine subspace ($=\Theta_a(Y)$ for any $a\in Y$).
Since I have not arrived at barycenter, I can't express elements in the spanned subspace using linear combination with sum of coefficients being 1. But this proposition appears before the concept of barycenter is introduced, so I think the equality can be proved without assistance of barycenter. I have proved $\overrightarrow{\langle M\cup N\rangle}\supseteq\overrightarrow{M}+\overrightarrow{N}+K\vec{ab}$. However, I just have no idea how to start proving $\subseteq$ part because the spanned subspace and its direction are too hard for me to handle. All I have at hand is the following two facts I have proven:
- $M\cap N\ne\emptyset$ iff $\overrightarrow{ab}\in\overrightarrow{M}+\overrightarrow{N}$ for any $a\in M$ and any $b\in N$.
- $M\cap N$ consists of a single point iff $\overrightarrow{ab}\in\overrightarrow{M}+\overrightarrow{N}$ for some $a\in M$ and some $b\in N$, and $\overrightarrow{M}\cap\overrightarrow{N}=\{0\}$.
But I don't know how to use them to establish the equality. It appears in two textbooks but both of them think it is too easy to provide a proof, leaving me so frustrated even after trying hard. Any help or hint would be greatly appreciated (not using barycenter). Thank you.
Got an idea while bathing, so I think I can prove the equality myself. Criticisms welcomed.
$\supseteq$:
Although we do not know the concrete form of elements in $\langle M\cup N\rangle$, we do know $M$ and $N$ are in it. Each element in $\overrightarrow{M}+\overrightarrow{N}+K\vec{ab}$ has the form $\overrightarrow{ax}+\overrightarrow{by}+\mu\overrightarrow{ab}$ where $x\in M,y\in N$ and $\mu\in K$. Taking $a$ as the origin, we map this element back to $X$, which is $\Theta_a(\overrightarrow{ax}+\overrightarrow{by}+\mu\overrightarrow{ab})=\Theta_a(\overrightarrow{ax})+\Theta_a(\overrightarrow{by})+\mu\Theta_a(\overrightarrow{ab})$, with the fact that $\Theta_a$ is a vector space isomorphism between $\vec X$ and vectorialization $X_a$. $\Theta_a(\overrightarrow{ax})=x\in M$, $\Theta_a(\overrightarrow{by})=\Theta_a(\overrightarrow{ba}+\overrightarrow{ay})=-b+y$. Note that the above operations are carried out in vector space $X_a$. Since $N$ is a vector subspace of $X_a$ which preserves values of all these operations, $-b+y\in N$. The third term $\mu\Theta_a(\overrightarrow{ab})=\mu b$ which is in $N$ too. As a result, these three terms are in either $M$ or $N$, thus in $\langle M\cup N\rangle$. Since $\langle M\cup N\rangle$ is a vector subspace of $X_a$, $x+(-b+y)+\mu b$ is still in this vector subspace, which leads to $\overrightarrow{ax}+\overrightarrow{by}+\mu\overrightarrow{ab}\in\Theta_a(\langle M\cup N\rangle)$ and in turn $\overrightarrow{\langle M\cup N\rangle}\supseteq\overrightarrow{M}+\overrightarrow{N}+K\vec{ab}$.
$\subseteq$:
Since $\langle M\cup N\rangle$ is the smallest affine subspace, i.e., smallest vector subspace in $X_a$, containing $M\cup N$, if we can prove that $\Theta_a^{-1}(\overrightarrow{M}+\overrightarrow{N}+K\vec{ab})$ which is a vector subspace already, contains $M\cup N$, we will have $\Theta_a^{-1}(\overrightarrow{M}+\overrightarrow{N}+K\vec{ab})$ contains $\langle M\cup N\rangle$ by definition of spanned subspace. From $\overrightarrow{M}\subseteq\overrightarrow{M}+\overrightarrow{N}+K\vec{ab}$, we have $M\subseteq\Theta_a^{-1}(\overrightarrow{M}+\overrightarrow{N}+K\vec{ab})$. Likewise we have $N\subseteq\Theta_b^{-1}(\overrightarrow{M}+\overrightarrow{N}+K\vec{ab})$. But $\Theta_a^{-1}(\overrightarrow{M}+\overrightarrow{N}+K\vec{ab})=\Theta_b^{-1}(\overrightarrow{M}+\overrightarrow{N}+K\vec{ab})$ because $a$ and $b$ are both in this inverse image. So, $\langle M\cup N\rangle\subseteq\Theta_a^{-1}(\overrightarrow{M}+\overrightarrow{N}+K\vec{ab})$, which completes the whole proof.