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I'm working through a question that I don't quite know how to get.

It is: Find an inner product $\langle - , -\rangle$ on $\mathbb{R} ^3$, and a matrix A such that $\langle u, v \rangle = u^\top Av$ such that, with respect to $\langle - , -\rangle$, the basis $\biggl \{(1, 1, 0)^\top , (1, 0, 1)^\top , (0, 1, 1)^\top\biggr\} $ is orthonormal.

So, normally if I had an inner product that I was looking to find the orthonormal basis for I'd use the Gram-Schmidt process and divide by the norms. However, it doesn't seem as if you can use that process in reverse. I'm not looking for an answer in any way, more of how I'd get started on working this through.

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Start by finding three vectors, each of which is orthogonal to two of the given basis vectors and then try and find a matrix $A$ which transforms each basis vector into the vector you've found orthogonal to the other two. This matrix gives you the inner product.

I would first work out the matrix representation $A'$ of the inner product with respect to the given basis, because then the columns of $A'$ are just the vectors that you've already found (since you want to transform each basis element into each of these vectors). It is important that you express the vectors you've found in terms of the basis given, and not the standard basis for this part.

Once you've done this, we can relate $A$ and $A'$ by $A = P^{\dagger}A'P$ where $P$ is the change of basis matrix from the given basis to the standard basis.

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  • $\begingroup$ Ok, so for instance, I'd find the vectors (1, -1, -1), (-1, -1, 1), and (-1, 1, -1) orthogonal to two basis vectors, and then find a transformation matrix A which converts one to the other? $\endgroup$
    – Jay
    Dec 9 '12 at 20:53
  • $\begingroup$ $A$ needs to convert $(0,1,1)$ to $(1,-1,-1)$ etc. You could do this the long way, or do it by computing $A'$ working in co-ordinates with respect to the given basis (so $(1,-1,-1)$ becomes $\frac{1}{2}(1,1,-2)$) and then use the change of basis matrix to convert back to the usual basis and get $A = P^{\dagger}A'P$. $\endgroup$ Dec 9 '12 at 21:02
  • $\begingroup$ @Thomas I've added to my answer to make what I was saying in the comment a little clearer, and hopefully provide a better way of answering the question. $\endgroup$ Dec 9 '12 at 21:15
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Let $u_1^\top = (1, 1, 0)^\top$ , $u_2^\top = (1, 0, 1)^\top $, $u_3^\top = (0, 1, 1)^\top $, and let $B$ the matrix $[u_1\ u_2\ u_3]$ and $A = (B B^\top)^{-1}$. $ u_i^\top A^{-1} u_j = (B^{-1}u_i )^\top B^{-1}u_j = e_i^\top e_j = \delta_{ij}.$ This generalizes to an arbitrary basis.

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