Finding an inner product such that these vectors are orthonormal

Question

The task goes like this:

Find an inner product in $\mathbb{R}^3$ such that the vectors $$a_1=(1,1,1),\qquad a_2=(1,0,1),\qquad a_3=(1,1,0)$$ are orthonormal.

How do I find the inner product?

Answer 1

Since $a_1=(1,1,1), a_2=(1,0,1), a_3=(1,1,0)$ is a basis. You can completely determine an inner product $\langle,\rangle$ by the values it takes on $a_1,a_2,a_3$. I can define the inner product $\langle,\rangle$ as $\langle a_1,a_2\rangle=0, \langle a_2,a_3\rangle=0, \langle a_3,a_1\rangle=0$ and $\langle a_1,a_1\rangle=1, \langle a_2,a_2\rangle=1, \langle a_3,a_3\rangle=1$.

In this case $\langle (x_1,x_2,x_3) , (y_1,y_2,y_3) \rangle = |x_2y_3+x_3y_2-x_1y_1|$ will do it.

Answer 2

Using properties of inner product, you can type $a\cdot b = a^TGb$ for symmetric matrix $G$. $G$ have 6 elements (because its symmetric). And we have system: $$a_i\cdot a_j=0,\enspace i\ne j,\quad a_i\cdot a_i = 1,\enspace i=1,2,3,$$ consists of 6 equations. Now we substitute your $a_i$'s, and find $$G = \begin{pmatrix}3 & -2 & -2\\-2 & 2 & 1\\-2 & 1 & 2\end{pmatrix},$$ and you have explicit formula for inner product.

Answer 3

An inner product $\langle\cdot,\cdot\rangle$ can be represented by a Euclidean inner product involving a symmetric positive definite matrix. Given linearly independent $a_i$, $i=1,\ldots,n$, we look for an SPD matrix $B$ such that $\langle a_i,a_j\rangle=a_i^TBa_j=\delta_{ij}$. If $A=[a_1,\ldots,a_n]$ (matrix with the $i$th column given by $a_i$), we have an equivalent condition $A^TBA=I$. Hence $B=A^{-T}A^{-1}=(AA^T)^{-1}$. Note that since $A$ is nonsingular, $B$ is indeed SPD. In this example, it gives $$ B=\left(\begin{array}{rrr}3&-2&-2\\-2&2&1\\-2&1&2\end{array}\right), $$ so $$ \langle x,y\rangle=3x_1y_1+2x_2y_2+2x_3y_3-2x_1y_2-2x_2y_1-2x_1y_3-2x_3y_1+x_2y_3+x_3y_2. $$