linear space W which contains three subspaces $V_1$, $V_2$, $V_3$ $⊂W$ such that $W=V_1⊕V_2=V_1⊕V_3=V_2⊕V_3$ Give an example of a linear space W of finite dimension $d \geq 4$ which contains three subspaces $V_1$, $V_2$, $V_3$ $⊂W$ such that $W=V_1⊕V_2=V_1⊕V_3=V_2⊕V_3$.
Is that possible? I mean the only possibility of $W=V_1⊕V_2=V_1⊕V_3=V_2⊕V_3$ that I can think of is $V_1=V_2=V_3$ but then $V_1⊕V_2=V_1⊕V_3=V_2⊕V_3 = \emptyset$. It's probably something basic and I wasted way to much time thinking of it.
 A: Does this work? Let $\{e_1,e_2,e_3,e_4\}$ be a basis for a 4-dimensional $V$, then take
$V_1=\langle e_1+e_2,e_1+e_3\rangle$
$V_2=\langle e_4+e_2,e_4-e_3\rangle$
$V_3=\langle e_1,e_4\rangle$.
It's not difficult to show that $V_i\oplus V_j=V$ for $i\neq j$. Also not difficult to show that $V_1\cap V_3=V_2\cap V_3=\emptyset$. The only difficult bit is to show that $V_1\cap V_2=\emptyset$, which I haven't done here.
A: If $V_1\oplus V_2 = W$ then $\dim V_1 +\dim V_2 = \dim W$. Similarly, $\dim V_2 +\dim V_3 = \dim V_3 +\dim V_1 = \dim W$, and we can conclude that $\dim V_1 =\dim V_2 =\dim V_3 = n$ and $\dim W = 2n$.
We can easily solve problem when $n=1$: we can take $V_1=\mathcal L\{(1,0)\}$, $V_2=\mathcal L\{(1,1)\}$ and $V_3=\mathcal L\{(0,1)\}$. You can easily see that $V_1\oplus V_2 = V_2 \oplus V_3 = V_3 \oplus V_1 = \mathbb R^2$.
Now we can solve problem for $n=2$, by taking $V'_1=V_1\times V_1$, $V'_2=V_2\times V_2$ and $V'_3=V_3\times V_3$. Then $V'_1\oplus V'_2 = V'_2 \oplus V'_3 = V'_3 \oplus V'_1 = \mathbb R^4$ holds.
A: Consider W =M2×3(R) with subspace V1,V2,V3 such that
V1={A€W:a11,a12&a13 are non zero and all other entries are 0s}
V2={A€W:a21,a22 & a23 are non-zero and all other entries are 0s}
V3={A€W: a23 ,a22& a13 are nonzero and all other entries are 0s}
We get dimV1=dimV2=dimV3=3
And their mutual intersection will give zero space.
