Show a $T:V \to V$ exists such that $T$ maps subspaces to each other, under a few general conditions Let $V = \mathbb{R^n}$ and let $U_1,U_2,W_1,W_2 \subset V$ be subspaces of $V$ of dimension $d$ such that $\dim(U_1 \cap W_1) = \dim(U_2 \cap W_2) = k,$ where $k \leq d \leq n.$
Prove that there exists a $T: V \to V$ such that $T(U_1) = U_2$ and $T(W_1) = W_2.$
Intuitively this makes sense, I've worked out some examples in $\mathbb{R^4}$ but the examples haven't helped me prove this in general so far.  Any hints/solutions?  In general, how does one approach the problem of "showing that a linear operator $T$ exists, given some conditions"?
 A: A simple way to construct a linear application is to define it on a basis.
For example you can take a basis $e_1,\dots , e_k$ of $U_1\cap W_1$. Then you can expand them to get bases $e_1,\dots,e_k,e_{k+1},\dots, e_{d}$ of $U_1$ and $e_1,\dots,e_k,e_{d+1},\dots, e_{2d-k}$ of $W_1$. The union of these two bases yields a family of linearly independent vectors $e_1,\dots,e_{2d-k}$; you can then expand it to get a basis $e_1,\dots,e_n$ of $\mathbb{R}^n$. You can also do the same with $U_2\cap W_2$, $U_2$ and $W_2$, and this gives you another basis $e_1',\dots,e_n'$.
Then there is a (unique) $T:\mathbb{R}^n\to \mathbb{R}^n$ such that $T(e_j)=e_j'$. Because of the way the bases were constructed, $T$ has the required properties.
A: Hints:


*

*We can refrain from $T$ being endo-morphism, i.e. let $T:V_1\to V_2$ (where now $V_1=V_2$), with $U_i,W_i\le V_i$ for $i=1,2$.

*Fix a basis of $U_i\cap W_i$ for both $i=1,2$

*Extend it on $U_i$, then on $W_i$ (adding $d-k$ basis vectors).

*Extend these for a basis of $V_i$.

*Map the given basis elements of $V_1$ to those of $V_2$.

