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On p.24 of the John M. Lee's Introduction to Smooth Manifolds (2nd ed.), he constructs the smooth structure of the Grassmannian. And when he tries to show Hausdorff condition, he says that for any 2 $k$-dimensional subspaces $P_1$, $P_2$ of $\mathbb{R}^n$, it is always possible to find a $(n-k)$-dimensional subspace whose intersection with both $P_1$ and $P_2$ are trivial.

My question: I think it is also intuitively obvious that we can always find $(n-k)$-dimensional subspace whose intersection with m subspaces $P_1,\ldots,P_m$ are trivial, which is a more generalized situation. But I can't prove it rigorously. Could you help me for this generalized case?

Thank you.

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There is no plane in $\Bbb{R}^3$ with trivial intersection with two planes that intersect only in a line. – Loki Clock Mar 30 '13 at 6:04
You just need a line whose intersection with the two planes is trivial. – copper.hat Mar 30 '13 at 6:05
If $P_i$ are planes, then $k=2$. If $n=3$, then $n-k=1$. – copper.hat Mar 30 '13 at 6:08
How the post is worded: Either it looks like 2k-dimensional and should be two k-dimensional, or "intersection with both" means intersection over the three subspaces. – Loki Clock Mar 30 '13 at 6:09
I don't see the ambiguity, sorry. – copper.hat Mar 30 '13 at 6:12
up vote 2 down vote accepted

If $n=k$, then $P_1=P_2=\mathbb{R}^n$ and you can take the zero subspace. So, we can assume there is a non-zero vector $e\in \mathbb{R}^n-P_1$.

Case 1: $e\notin P_2$. Set $e_1:=e$.

Case 2: $e\in P_2$. Find $x\in P_1$ such that $x\notin P_2$, then $e+x\notin P_1$ and $e+x\notin P_2$. Set $e_1:=e+x$.

It is clear that $P_1\oplus \langle e_1\rangle$ and $P_2\oplus \langle e_1 \rangle$ are two $k+1$-dimensional subspaces of $\mathbb{R}^n$. If $P_1\oplus \langle e_1\rangle\neq \mathbb{R}^n$, you can repeat this process for these two subspaces to find $e_2$.

By induction, you find $e_1,\cdots, e_{n-k}$ such that they are independent and non of them belong to $P_1$ or $P_2$. The subspace generated by $e_1,\cdots, e_{n-k}$ is your answer.

Edit: In order to address the issue raised by @GeorgesElencwajg , we have to note that not only $e_1,\cdots,e_{n-k}$ are linearly independent, but also no non-trivial linear combinations of $e_1,\cdots,e_{n-k}$ belongs to $P_i$ for $i=1,2$. And this claim can be easily verified regarding our construction.

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The last part of your answer is false: in $\mathbb R^3$ with basis $v_1,v_2,v_3$ take $P_1=\mathbb R\cdot v_1,\:P_2=\mathbb R\cdot v_2,\: e_1=v_1+v_2, \:e_2=v_1-v_2$. The subsapace generated by $e_1, e_2$ does not have trivial intersection with $P_1$ nor $P_2$ although $e_1,e_2$ are independent and do not belong to $P_1$ nor $P_2$. – Georges Elencwajg Mar 30 '13 at 7:43
@GeorgesElencwajg: You are right the argument was not complete. But I think the revised argument is okay now. – Vahid Shirbisheh Mar 30 '13 at 8:34

For $j=1,2,\ldots,m$, let $B_j$ be a basis of $P_j$. It suffices to find a set of vectors $S=\{v_1,v_2,\ldots,v_{n-k}\}$ such that $S\cup B_j$ is a linearly independent set of vectors for each $j$. We will begin with $S=\phi$ and put vectors into $S$ one by one.

Suppose $S$ already contains $i<n-k$ vectors. Now consider a vector $v_{i+1}=v_{i+1}(x)$ of the form $(1,x,x^2,\ldots,x^{n-1})^T$. For each $j$, there are at most $n-1$ different values of $x$ such that $v_{i+1}(x)\in\operatorname{span}\left(\{v_1,v_2,\ldots,v_i\}\cup B_j\right)$, otherwise there would exist a non-invertible Vandermonde matrix that corresponds to distinct interpolation nodes. Therefore, there exist some $x$ such that $v_{i+1}(x)\notin\operatorname{span}\left(\{v_1,v_2,\ldots,v_i\}\cup B_j\right)$ for all $j$. Put this $v_{i+1}$ into $S$, $S\cup B_j$ is a linearly independent set. Continue in this manner until $S$ contains $n-k$ vectors. The resulting $\operatorname{span}(S)$ is the subspace we desire.

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