Section 6.2 Gram-Schmidt Orthogonalization Process
In this section, we look at how to construct an orthogonal basis of \(\R^n\) from a basis.
Let \(\{v_1,\ldots, v_n\}\) be a basis of \(\R^n\text{.}\) Define
\begin{align*}
u_1: =\amp v_1\\
u_2: =\amp v_2 -\frac{u_2\cdot u_1}{\norm{u_1}^2}u_1\\
u_3: =\amp v_3 -\frac{v_3\cdot u_1}{\norm{u_1}^2}u_1-\frac{v_3\cdot u_2}{\norm{u_2}^2}u_2\\
\amp \vdots\\
u_n: =\amp v_n -\frac{v_n\cdot u_1}{\norm{u_1}^2}u_1-\cdots -\frac{v_n\cdot u_{n-1}}{\norm{u_{n-1}}^2}u_{n-1}
\end{align*}
In view of Ex.
Checkpoint 6.1.2, it is easy to see that
\(\{u_1,\ldots,
u_n\}\) is an orthogonal basis of
\(\R^n\text{.}\) Now we normalize
\(u_i's\text{.}\) Define
\(q_i=\frac{u_i}{\norm{u_i}}\text{.}\) Then
\(\{q_1,\cdots, q_n\}\) is an orthononal basis of
\(\R^n\text{.}\) Note that we could have defined
\(q_i\) immediately after defining
\(u_i\text{.}\)
This process is called the Gram-Schmidt orthogonalization process.
Geometrically \(u_2\text{,}\) constructed by subtracting the orthogonal projection of \(v_2\) on to \(u_1\text{.}\) In order to construct \(u_3\text{,}\) we take sum of orthogonal projections of \(v_3\) onto \(u_1\) and \(u_2\text{,}\) which is the orthogonal projection of the plane spanned by \(u_1\) and \(u_2\) and subtract this from \(v_3\text{.}\) Readers are encouraged to draw figures.
Example 6.2.1.
Use the Gram-Schmidt orthogonalization process to find an orthonormal basis of \(\R^3\) starting with a basis \(\{(0,1,1), (1,1,1), (1,-2,2)\}\text{.}\) Let \(v_1 = (0, 1, 1), v_2=(1, 1, 1), v_3=(1, -2, 2)\text{.}\) Then we have
\begin{align*}
u_1: = \amp (0, 1, 1)\\
u_2: = \amp v_2 -\frac{u_2\cdot u_1}{\norm{u_1}^2}u_1=(1, 1, 1)-\frac{2}{2}(0,1,1)=(1,0,0)\\
u_3:= \amp v_3 -\frac{u_3\cdot u_1}{\norm{u_1}^2}u_1-\frac{u_3\cdot u_2}{\norm{u_2}^2}u_2\\
= \amp (1, -2, 2)-\frac{0}{2}(0,1,1)-\frac{1}{1}(1,0,0)=(0,-2,2)
\end{align*}
Thus the orthonormal basis obtained from the given basis is
\begin{equation*}
\left\{q_1 = \left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right), q_2=(1,0,0), q_3 =\left(0,\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\right\}\text{.}
\end{equation*}
Example 6.2.2.
Consider the matrix \(A=\left(\begin{array}{rrrr} -1 \amp 1 \amp 0 \amp 1 \\ 1 \amp -1 \amp 0 \amp 1 \\ -1 \amp 0 \amp 2 \amp 1 \end{array} \right)\text{.}\) Find an orthogonal basis of the row space of \(A\text{.}\) It is easy to check that rank of \(A\) is 3. Hence row are linearly independent vectors in \(\R^4\text{.}\)
Let \(v_1=(-1,1,0,1), v_2 = (1,-1,0,1), v_3=(-1,0,2,1)
\text{.}\)
\begin{align*}
u_1: = \amp (-1,1,0,1)\\
u_2: = \amp v_2 -\frac{u_2\cdot u_1}{\norm{u_1}^2}u_1\\
=\amp (1,-1,0,1)-\frac{-1}{3}(-1,1,0,1)=(2/3, -2/3, 0, 4/3)\\
u_3: = \amp v_3 -\frac{u_3\cdot u_1}{\norm{u_1}^2}u_1-\frac{u_3\cdot u_2}{\norm{u_2}^2}u_2\\
= \amp (-1,0,2,1)-\frac{2}{3}(-1,1,0,1)-\frac{2/3}{9/2}(2/3, -2/3, 0, 4/3)\\
=\amp (-1/2, -1/2, 2, 0)
\end{align*}
Hence
\begin{equation*}
\{(-1, 1, 0, 1) (2/3, -2/3, 0, 4/3) (-1/2, -1/2, 2, 0)\}
\end{equation*}
is an orthogonal basis of the row space of \(A\text{.}\)
Checkpoint 6.2.3.
Use the Gram-Schmidt orthogonalization process to find an orthonormal basis of \(\R^3\) starting with a basis \(\beta=\{(1,1,1),(-1,1,1),(-1,0,1)\}\text{.}\)
Checkpoint 6.2.4.
Use the Gram-Schmidt orthogonalization process to find an orthonormal basis of the subspace \(W\subset \R^4\) with basis
\begin{equation*}
\beta = \{ (-1,1,1,0),(-1,0,1,0),(1,0,0,1)\}\text{.}
\end{equation*}
