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

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 $$W\oplus V=K$$ 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.

• You say that you want to avoid $\mathbb R^n$, but your spaces are clearly subspaces of $\mathbb R^4$. – lulu Jul 23 '20 at 15:28
• Your question is not clear. Unless I made an arithmetic error, the vector $B_V$ is not in the subspace of $\mathbb R^4$ spanned by the two vectors you wrote out. – lulu Jul 23 '20 at 15:37
• Sorry, what I meant by that was that the examples I have seen were looking for two complementary subspaces V, W such that W (+) V = R^n; in this problem, W (+) V do not sum to R^n, but to K – Jon G Jul 23 '20 at 15:38
• But, as I say, unless I blundered $V$ is not a subspace of $K$. – lulu Jul 23 '20 at 15:51
• @lulu, you were correct. I was calculating vector $B_V$ incorrectly. I have updated the question to include the correct vector, and this time I made sure it was in $K$ :) – Jon G Jul 23 '20 at 16:27

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.
• I understand the process you're describing, but just a question on finding a complementary vector to $(4/5,2/5)$ in $R^2$. I think what you're indicating is that the dot product of $(4/5,2/5)$ and the complementary vector should be 0, is that correct? – Jon G Jul 23 '20 at 23:06