Let $M_1$ and $M_2$ be affine subspaces of $\Bbb R^n$ with $M_1 \cap M_2 \neq \emptyset$ and $\dim(M_1 \cap M_2)=\dim(M_1)$. Then $M_1 \subset M_2$.

Question

Let $M_1$ and $M_2$ be affine subspaces of $\mathbb{R}^n$ with $M_1 \cap M_2 \neq \emptyset$ and $\dim(M_1 \cap M_2)=\dim(M_1)$. Then $M_1 \subset M_2$. I tried to prove by contradiction and took an element in $M_1$ and assumed it was not in the intersection, but it did not work. I also tried to use the dimensions of linear subspaces to create these affine subspaces, but it did not work. Any help is appreciated.

Answer 1

Let $a\in M_1\cap M_2$

Let $U_1$ and $U_2$ two linear subspaces of $\mathbb R^n$ such that $$M_i=a+U_i(i=1,2)$$

By definition, $\dim M_i=\dim U_i$.

On the other hand, we obviously have that$$M_1\cap M_2=a+U_1\cap U_2$$

So, by hypothesis, we have that $$\dim U_1\cap U_2=\dim U_1$$

$U_1\cap U_2$ is a linear subspace of $U_1$. Therefore $$U_1\cap U_2=U_1$$

So$$M_1=a+U_1=a+U_1\cap U_2\subset M_2=a+U_2.\square$$

Answer 2

Without loss of generality assume that $M_1$ and $M_2$ are linear subspaces of $\mathbb{R}^n$. Pick a basis of $M_1 \cap M_2$, say $v_1, \ldots, v_k$. But then $v_1, \ldots, v_k$ lie in particular in $M_1$, so by the dimensionality assumption they span $M_1$ and are therefore a basis, whence $M_1 \subseteq M_2$.