Finding a subspace whose intersections with other subpaces are trivial. 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.
 A: 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.  
A: 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.
