0
$\begingroup$

Let $W$ be a subspace of the complex vector space $\mathbb{C}^4$. Given that the inner product on $\mathbb{C}^4$ is defined as $$\langle a,b\rangle=a_1\bar{b_1}+a_2\bar{b_2}+a_3\bar{b_3}+a_4\bar{b_4},$$ and the basis for $W$ is $$\mathcal{B}=\{(1,0,-1,0),(0,1,0,i)\},$$ find the basis for the orthogonal complement $W_{\perp}$.

I was wondering if there is a quick way to solve a problem of this kind, or do I have to define arbitrary vectors $a,b\in W_{\perp}$ such that each vector is orthogonal to each vector in $\mathcal{B}$ and are themselves linearly independent?

$\endgroup$
  • 1
    $\begingroup$ This particular problem seems to want you to just choose two other vectors in a clever way. In particular, a vector which is zero in its first and third components will immediately be orthogonal to your first vector, so choose its second and fourth components to make the dot product with the second vector vanish. A similar trick (making the second and fourth components zero) will give you a second vector. $\endgroup$ – 211792 Oct 29 '15 at 5:16
0
$\begingroup$

Let u=(x,y,z,w) be a vector of the orthogonal complement of W, where W is the span of the given two vectors. Then, the following have to be satisfied. 1) <(1,0,-1,0), (x,y,z,w)>=0 2) <(0,1,0,i), (x,y,z,w)>=0 Essentially, these conditions are the same as the following conditions. 1) iff <(x,y,z,w), (1,0,-1,0)>=0 2) iff <(x,y,z,w), (0,1,0,i)>=0 Then, by 1), we have x-z=0, and by 2), we have y=iw. Reflecting this result, we can rewrite u=(x,iw,x,w)=x(1,0,1,0)+w(0,i,0,1). Therefore, the orthogonal complement of W=span{(1,0,1,0),(0,i,0,1)}.

$\endgroup$
  • 2
    $\begingroup$ Please consider using MathJax to make your answer more legible. $\endgroup$ – Roland Mar 16 '16 at 6:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy