algebra - scalar product and ortonormal bases

Question

I've got an ALGEBRA question.

We've got the scalar product · defined in R3 and the base B={(1,0,-1),(1,-1,-1),(0,1,1)} is orthonormal to that scalar product. We have to calculate the general expression of (x,y,z)·(x',y',z')

This is what I have done:

As I know that B is orthonormal respect that scalar product, I know that when I do the scalar product of each vector with its own is 1 and the scalar product of two different vectors is 0.

So I've proposed 6 diferent equations. Like this, I've obtain the following matrix:

First line of the matrix: (2,-1,2)

Second line of the matrix:(-1,2,-2)

Third line of the matrix: (2,-2,3)

If I'm not wrong that's OK but how can I solve this problem in a shorter or easier way?

Answer 1

Let $$B=\begin{bmatrix}1&1&0\\0&-1&1\\-1&-1&1\end{bmatrix},$$ i.e., the matrix with the elements of $B$ as its columns. Since this basis is orthonormal with respect to the inner product, your problem is equivalent to finding a matrix $Q$ such that $B^TQB=I$. Multiply this equation by the appropriate inverses to isolate $Q$.

Answer 2

Let $v_1 = (1,0,-1)$, $v_2 = (1,-1,-1)$ and $v_3 = (0,1,1)$. It is straightforward to check that $$(x,y,z) = (y-z)v_1 + (x-y+z)v_2 + (x+z)v_3$$ for any $(x,y,z) \in \mathbb{R}^3$. Then, using the linearity of "$\cdot$" and the fact that $v_i \cdot v_j = 1$ if $i=j$ and $0$ in another case, we have that \begin{align} (x,y,z) \cdot (x',y',z') = (y-z)(y'-z') + (x-y+z)(x'-y'+z') + (x+z)(x'+z'). \end{align}