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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.

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    $\begingroup$ You say that you want to avoid $\mathbb R^n$, but your spaces are clearly subspaces of $\mathbb R^4$. $\endgroup$ – lulu Jul 23 '20 at 15:28
  • $\begingroup$ 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. $\endgroup$ – lulu Jul 23 '20 at 15:37
  • $\begingroup$ 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 $\endgroup$ – Jon G Jul 23 '20 at 15:38
  • $\begingroup$ But, as I say, unless I blundered $V$ is not a subspace of $K$. $\endgroup$ – lulu Jul 23 '20 at 15:51
  • $\begingroup$ @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 $K$ :) $\endgroup$ – Jon G Jul 23 '20 at 16:27
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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.

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  • $\begingroup$ 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? $\endgroup$ – Jon G Jul 23 '20 at 23:06
  • $\begingroup$ @JonG Yes, it’s one of the ways to find a complement, in this case the orthogonal complement with respect to the standard inner product. $\endgroup$ – egreg Jul 24 '20 at 8:45

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