# How to find orthonormal basis for inner product space?

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?

• Maybe you should be more specific about your problem. You said it yourself, we use Gram-Schmidt. It is just an algorithm. An easier way to get your basis is to see that $((1,0,0),(0,1,0),(0,1,0))$ is still an orthogonal basis. Just re-scale those vectors. – Ivo Terek Jul 30 '14 at 10:45
• Do you know all the steps of the Gram-Schmidt process? Did you try them (and can you show them here)? – dreamer Jul 30 '14 at 10:54
• According to @Ivo Terek comment, the solution is $(1,0,0)$, $\frac{1}{\sqrt{2}}(0,1,0)$ and $\frac{1}{\sqrt{3}}(0,0,1)$ are the solution. – Mojtaba Golshani Jul 30 '14 at 11:10

Three steps which will always result in an orthonormal basis for $\mathbb R^n$:

1. Take a basis $\{w_1,w_2,\dots,w_n\}$ for $\mathbb R^n$ (any basis is good)
2. Orthogonalize the basis (using gramm-schmidt), resulting in a orthogonal basis $\{v_1,v_2,\dots,v_n\}$ for $\mathbb R^n$
3. 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 $a,b,c$.

• u_{2}, u_{3} has a norm 1 in your answer? – Noor Aslam Apr 17 '18 at 11:56