example of Gram-Schmidt orthogonalization

Let us work with the standard inner product on $\\mathbbmss{R}^{3}$ (dot product) so we can get a nice geometrical visualization.

Consider the three vectors

	$\\displaystyle v_{1}$	$\\displaystyle=(3,0,4)$
	$\\displaystyle v_{2}$	$\\displaystyle=(-6,-4,1)$
	$\\displaystyle v_{3}$	$\\displaystyle=(5,0,-3)$

which are linearly independent (the determinant of the matrix $A=(v_{1}|v_{2}|v_{3})=116\eq 0)$ but are not orthogonal.

We will now apply Gram-Schmidt to get three vectors $w_{1},w_{2},w_{3}$ which span the same subspace (in this case, all $R^{3}$ ) and orthogonal to each other.

First we take $w_{1}=v_{1}=(3,0,4)$ . Now,

w_{2}=v_{2}-\\frac{w_{1}\\cdot v_{2}}{\\|w_{1}\\|^{2}}w_{1}

that is,

w_{2}=(\\frac{-108}{25},-4,\\frac{81}{25})

and finally

w_{3}=v_{3}-\\frac{w_{1}\\cdot v_{3}}{\\|w_{1}\\|^{2}}w_{1}-\\frac{w_{2}\\cdot v_{3}%}{\\|w_{2}\\|^{2}}w_{2}

which gives

w_{3}=(\\frac{1856}{1129},\\frac{3132}{1129},\\frac{1392}{1129})

and so $\\{w_{1},w_{2},w_{3}\\}$ is an orthogonal set of vectors such that $\\langle w_{1},w_{2},w_{3}\\rangle=\\langle v_{1},v_{2},v_{3}\\rangle$ .

If we rather consider $\\{w_{1}/\\|w_{1}\\|,w_{2}/\\|w_{2}\\|,w_{3}/\\|w_{3}\\|\\}$ then we get an orthonormal set.