In $\mathbb{R}^3$ we declare an inner product as follows: $\langle v,u \rangle \:=\:v^t\begin{pmatrix}1 & 0 & 0 \\0 & 2 & 0 \\0 & 0 & 3\end{pmatrix}u$
How can I find an orthonormal basis for this inner product space using the Gram–Schmidt process?
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$\begingroup$Three steps which will always result in an orthonormal basis for $\mathbb R^n$:
- Take a basis $\{w_1,w_2,\dots,w_n\}$ for $\mathbb R^n$ (any basis is good)
- Orthogonalize the basis (using gramm-schmidt), resulting in a orthogonal basis $\{v_1,v_2,\dots,v_n\}$ for $\mathbb R^n$
- Normalize the vectors $v_i$ to obtain $u_i=\frac{v_i}{||v_i||}$ which form a orthonormal basis.
First you should know that orthonormal means "orthogonal plus the vectors have length $1$. The following is an orthonormal basis for the given inner product
$$ \left\{ u_1=(1,0,0),u_2=\left( 0,\frac{1}{\sqrt{2}},0 \right), u_3=\left(0,0,\frac{1}{\sqrt{3}}\right) \right\}. $$
You can check that the vectors are othogonal and have length of unity. To find them assume that they have the forms respectively
$$ u_1=(a,0,0),u_2=(0,b,0), u_3 = (0,0,c) $$
then use the definition of the inner product you have been given to find $a,b,c$.
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