How to find a basis of complementary subspace of a subspace not in $mathbb R^n$?

Question

My question is similar to this question, but I am trying to find a complementary subspace of a subspace that is not in $mathbb R^n$.
I am trying to find a subspace $W$ with basis vector $B_W$ such that $$Woplus V=K$$ where $K$ is a subspace with basis vectors $$B_K=left{(1.5, -0.5, 1.5, -1), (-0.5, 3.5, -0.5, -3)right}$$ and $V$ is a subspace with basis vectors $$B_V=left{(1, 1, 1, -2)right}.$$
Based on the definition of a complementary subspace I found here and the method for finding the intersection of two subspaces that I found here, I believe I can set up this system of equations: $$text{span}(B_K)=text{span}(B_V)+text{span}(B_W)$$$$text{span}(B_V)=text{span}(B_W).$$
Substituting in the vectors, I have:$$a_1(1.5, -0.5, 1.5, -1)+a_2(-0.5, 3.5, -0.5, -3)=b_1(1,1,1,-2)+b_2(B_W)$$$$c_1(1,1,1,-2)=c_2(B_W).$$
At this point I'm not sure how to solve this system to find $B_W$. Any help in solving this system - or if there is another method which would be more helpful - would be greatly appreciated.

egreg · Answer

You can use what you know about $mathbb{R}^n$. Let $v_1=(3/2,-1/2,3/2,-1)$ and $v_2=(-1/2,7/2,-1/2,-3)$.
The coordinates of $v=(1,1,1,-2)$ with respect to the basis ${v_1,v_2}$ are $(4/5,2/5)$.
Now you can find a complementary vector of $(4/5,2/5)$ in $mathbb{R}^2$, for instance $(2,-4)$ and your needed vector will be
$$
2v_1-4v_2=(3,-1,3,-2)-(-2,14,-2,-12)=(5,15,5,10)
$$
Of course it is not unique. You could as well use $v_1$ or $v_2$. However this method extends to any dimension.

How to find a basis of complementary subspace of a subspace not in $mathbb R^n$?

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