Answer to Question #162363 in Linear Algebra for Nikhil

Question

Complete {(2,0,3)} to form an orthogonal basis of R^3

Answer 1

Solution:

According to the Gram-Schmidt process,

$\mathbf{u}_{\mathbf{k}}=\mathbf{v}_{\mathbf{k}}-\sum_{j=1}^{k-1} \operatorname{proj}_{\mathbf{u}_{\mathbf{j}}}\left(\mathbf{v}_{\mathbf{k}}\right)$ , where

$\operatorname{proj}_{\mathbf{u}}(\mathbf{v})=\dfrac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u}$

The normalized vector is 

$\mathbf{e}_{\mathbf{k}}=\dfrac{\mathbf{u}_{\mathbf{k}}}{\sqrt{\mathbf{u}_{\mathbf{k}} \cdot \mathbf{u}_{\mathbf{k}}}}$

Here $\cdot$ is the dot product operator

$\mathbf{u}_{\mathbf{1}}=\mathbf{v}_{\mathbf{1}}=\left[\begin{array}{l}2 \\ 0 \\ 3\end{array}\right]$

Now, $\mathbf{e}_{1}=\dfrac{\mathbf{u}_{1}}{\sqrt{\mathbf{u}_{1} \cdot \mathbf{u}_{1}}}$

$\sqrt{\mathbf{u}_{1} \cdot \mathbf{u}_{1}}=\sqrt{2^2+0^2+3^2}=\sqrt{4+9}=\sqrt{13}$

So, $\mathbf{e}_{1}=\dfrac{\mathbf{u}_{1}}{\sqrt{\mathbf{u}_{1} \cdot \mathbf{u}_{1}}}=\left[\begin{array}{c}\dfrac{2 \sqrt{13}}{13} \\ 0 \\ \dfrac{3 \sqrt{13}}{13}\end{array}\right]$

So, the required set of orthonormal vectors is $\mathbf{e}_{1}=\left[\begin{array}{c}\dfrac{2 \sqrt{13}}{13} \\ 0 \\ \dfrac{3 \sqrt{13}}{13}\end{array}\right]$