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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – Ivo Terek
    Jul 30, 2014 at 10:45
  • $\begingroup$ Do you know all the steps of the Gram-Schmidt process? Did you try them (and can you show them here)? $\endgroup$
    – dreamer
    Jul 30, 2014 at 10:54
  • $\begingroup$ 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. $\endgroup$ Jul 30, 2014 at 11:10

2 Answers 2


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 $a,b,c$.

  • $\begingroup$ u_{2}, u_{3} has a norm 1 in your answer? $\endgroup$
    – Noor Aslam
    Apr 17, 2018 at 11:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.