Decomposition of vector space by intersections

Question

If a vector space has two deompositions $V = V_1 \oplus V_2 = W_1 \oplus W_2$, then is it true that $V = \oplus_{i,j=1,2} V_i \cap W_j$

Of course, I guess so. But the following example is not so. Probably I have misunderstand but I cannot found it. Please let me know where is wrong.

Let $V$ be a three dimensional vector space and denote its basis by $e_1, e_2,e_3$.

Take another basis $f_1,f_2,f_3$ constructed from $e_1, e_2,e_3$ as follows. $$\begin{align*} f_1 &= e_1 +e_2,\\ f_2 &= e_3,\\ f_3 &= e_2 +e_3. \end{align*}$$

This transformation matrix has non-zero determinant.

Consider two decompositions of V, namely,

$V= (e_1,e_2) \oplus (e_3)=: V_1 \oplus V_2,$

$V= (f_1,f_2) \oplus (f_3)=: W_1 \oplus W_2.$

From the following calculations

$V_1 \cap W_1 = (f_1),$

$V_1 \cap W_2 = 0,$

$V_2 \cap W_1 = (e_3),$

$V_2 \cap W_2 = 0,$

I cannot deduce $V = \oplus_{i,j=1,2} V_i \cap W_j$

Answer 1

Indeed your desired formula does not hold. Consider for example the two bases $\{(1,0),(0,1)\}$ and $\{(1,0),(1,1)\}$ of $K^2$, where $K$ denotes some field. Then $$\begin{align*} K\cdot (1,0) \cap K\cdot (1,0) &= K \cdot (1,0)\\ K\cdot (1,0) \cap K\cdot (1,1) &= \{0\}\\ K \cdot (0,1) \cap K \cdot (1,0) &= \{0\}\\ K \cdot (0,1) \cap K \cdot (1,1) &= \{0\} \end{align*}$$ cannot possibly sum up to form $K^2$...

This proves in particular, that any vector space of dimension $\geq 2$ admits two such decompositions, which cannot be refined to make your formula hold.