# The Gram Schmidt Process for Orthonormal Basis

## The Gram Schmidt Process and Formulas

The Gram Schmidt process is used to produce an Orthonormal Basis for a subspace.
Given a basis for subspace V , the basis where

is an orthogonal basis for the subspace V .

The orthonormal basis Y0 is obtained by dividing each vector in the basis Y by its norm.

   

## Examples with Solutions

Example 1
Subspace $$V$$ is defined by span$$\{\textbf{v}_1 , \textbf{v}_2 \}$$ where $$\textbf{v}_1$$ and $$\textbf{v}_2$$ are column vectors given by
$$\textbf{v}_1 = \begin{bmatrix} 1\\ 2 \\ 0 \end{bmatrix}$$ and $$\textbf{v}_2 = \begin{bmatrix} -2 \\ -4 \\ -4 \end{bmatrix}$$
Use the Gram Schmidt process defined above to determine an orthonormal basis $$Y_O$$ for $$V$$

Solution to Example 1
Let $$Y = \{ \textbf{y}_1 , \textbf{y}_2 \}$$ be the orthogonal basis to determine. According to the fomrmulas above, we write
$$\textbf {y}_1 = \textbf {v}_1 = \begin{bmatrix} 1\\ 2 \\ 0 \end{bmatrix}$$

$$\textbf {y}_2 = \textbf {v}_2 - \dfrac{\textbf {v}_2 \cdot \textbf {y}_1}{\textbf {y}_1 \cdot \textbf {y}_1} \textbf {y}_1$$
Evaluate the inner product in the numerator and denominator
$$\dfrac{\textbf {v}_2 \cdot \textbf {y}_1}{\textbf {y}_1 \cdot \textbf {y}_1} = \dfrac{\begin{bmatrix} -2 \\ -4 \\ -4 \end{bmatrix} \cdot \begin{bmatrix} 1\\ 2 \\ 0 \end{bmatrix} }{ \begin{bmatrix} 1\\ 2 \\ 0 \end{bmatrix} \cdot \begin{bmatrix} 1\\ 2 \\ 0 \end{bmatrix} } = \dfrac{-10}{5} = -2$$
Substitute the above and evaluate $$\textbf {y}_2$$
$$\textbf {y}_2 = \begin{bmatrix} -2 \\ -4 \\ -4 \end{bmatrix} - (-2) \begin{bmatrix} 1\\ 2 \\ 0 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \\ -4 \end{bmatrix}$$

Hence $$Y = \left \{\begin{bmatrix} 1\\ 2 \\ 0 \end{bmatrix} , \begin{bmatrix} 0\\ 0 \\ -4 \end{bmatrix} \right \}$$
The orthonormal basis $$Y_O$$ is obtained by dividing each vector in the basis $$Y$$ by its norm.
$$Y_O = \left \{ \dfrac{1}{\sqrt 3}\begin{bmatrix} 1\\ 2 \\ 0 \end{bmatrix} , \dfrac{1}{4} \begin{bmatrix} 0\\ 0 \\ -4 \end{bmatrix} \right \}$$

Note that
1) $$Y = \{ \textbf {y}_1 , \textbf {y}_2 \}$$ is a basis for $$V$$ because it is a linear combination of $$\textbf {v}_1$$ and $$\textbf {v}_2$$.
We know that $$y_1 = v_1$$ and it can easily be shown that $$y_2 = 2 v_1 + v_2$$
Hence span $$\{ \textbf {v}_1, \textbf{v}_2 \}$$ = span $$\{ \textbf {y}_1, \textbf{y}_2 \}$$
2) the inner product of $$\textbf {y}_1$$ and $$\textbf {y}_2$$ given by $$\begin{bmatrix} 1\\ 2 \\ 0 \end{bmatrix} \cdot \begin{bmatrix} 0\\ 0 \\ -4 \end{bmatrix} = 0$$ means that $$\textbf {y}_1$$ and $$\textbf {y}_2$$ are orthogonal and hence $$Y$$ is an orthogonal basis for $$V$$

Example 2
Subspace $$V$$ is defined by span$$\{\textbf{v}_1 , \textbf{v}_2 , \textbf{v}_3 , \textbf{v}_4\}$$ where
$$\textbf{v}_1 = \begin{bmatrix} -1\\ -1 \\ 2 \\ 2\\ 3 \end{bmatrix}$$ , $$\textbf{v}_2 = \begin{bmatrix} 1 \\ 1 \\ 1\\ 1\\ 0 \end{bmatrix}$$ , $$\textbf{v}_3 = \begin{bmatrix} 0 \\ 0 \\ -1\\ -1\\ 0 \end{bmatrix}$$ , $$\textbf{v}_4 = \begin{bmatrix} 0 \\ 0 \\ 0\\ 1\\ 1 \end{bmatrix}$$
Use the Gram Schmidt process defined above to determine an
orthonormal basis $$Y_O$$ for $$V$$

Solution to Example 2
Let $$Y = \{\textbf{y}_1,\textbf{y}_2,\textbf{y}_3,\textbf{y}_4\}$$ be the
orthogonal basis to determine. Using the fomrmulas above, we write
$$\textbf {y}_1 = \textbf {v}_1 = \begin{bmatrix} -1\\ -1 \\ 2 \\ 2\\ 3 \end{bmatrix}$$

$$\textbf {y}_2 = \textbf {v}_2 - \dfrac{\textbf {v}_2 \cdot \textbf {y}_1}{\textbf {y}_1 \cdot \textbf {y}_1} \textbf {y}_1$$
Evaluate the inner product in the numerator and denominator
$$\dfrac{\textbf {v}_2 \cdot \textbf {y}_1}{\textbf {y}_1 \cdot \textbf {y}_1} = \dfrac{\begin{bmatrix} 1 \\ 1 \\ 1\\ 1\\ 0 \end{bmatrix} \cdot \begin{bmatrix} -1\\ -1 \\ 2 \\ 2\\ 3 \end{bmatrix} }{ \begin{bmatrix} -1\\ -1 \\ 2 \\ 2\\ 3 \end{bmatrix} \cdot \begin{bmatrix} -1\\ -1 \\ 2 \\ 2\\ 3 \end{bmatrix} } = \dfrac{2}{19}$$
Substitute the above and evaluate $$\textbf {y}_2$$
$$\textbf {y}_2 = \begin{bmatrix} 1 \\ 1 \\ 1\\ 1\\ 0 \end{bmatrix} - \dfrac{2}{19} \begin{bmatrix} -1\\ -1 \\ 2 \\ 2\\ 3 \end{bmatrix} = \begin{bmatrix} \frac{21}{19}\\ \frac{21}{19}\\ \frac{15}{19}\\ \frac{15}{19}\\ -\frac{6}{19}\end{bmatrix}$$
Note that if we mutliply $$\textbf{y}_2$$ by any number not equal to zero, it would not change the basis. Hence multiply $$\textbf{y}_2$$ by $$\dfrac{19}{3}$$ and simplify to
$$\textbf{y'}_2 = \dfrac{19}{3} \begin{bmatrix} \frac{21}{19}\\ \frac{21}{19}\\ \frac{15}{19}\\ \frac{15}{19}\\ -\frac{6}{19}\end{bmatrix} = \begin{bmatrix} 7\\ 7\\ 5\\ 5\\ -2 \end{bmatrix}$$
We now use vector $$\textbf {y'}_2$$ instead of $$\textbf {y}_2$$ in the formulas for $$\textbf {y}_3$$.

$$\textbf {y}_3 = \textbf {v}_3 - \dfrac{\textbf {v}_3 \cdot \textbf {y}_1}{\textbf {y}_1 \cdot \textbf {y}_1} \textbf {y}_1 - \dfrac{\textbf {v}_3 \cdot \textbf {y'}_2}{\textbf {y'}_2 \cdot \textbf {y'}_2} \textbf {y'}_2$$
Substitute
$$\textbf {y}_3 = \begin{bmatrix} 0 \\ 0 \\ -1\\ -1\\ 0 \end{bmatrix} - \dfrac{\begin{bmatrix} 0 \\ 0 \\ -1\\ -1\\ 0 \end{bmatrix} \cdot \begin{bmatrix} -1\\ -1 \\ 2 \\ 2\\ 3 \end{bmatrix} }{\begin{bmatrix} -1\\ -1 \\ 2 \\ 2\\ 3 \end{bmatrix} \cdot \begin{bmatrix} -1\\ -1 \\ 2 \\ 2\\ 3 \end{bmatrix} } \begin{bmatrix} -1\\ -1 \\ 2 \\ 2\\ 3 \end{bmatrix}$$

$$- \dfrac{\begin{bmatrix} 0 \\ 0 \\ -1\\ -1\\ 0 \end{bmatrix} \cdot \begin{bmatrix} 7\\ 7\\ 5\\ 5\\ -2 \end{bmatrix} }{\begin{bmatrix} 7\\ 7\\ 5\\ 5\\ -2 \end{bmatrix} \cdot \begin{bmatrix} 7\\ 7\\ 5\\ 5\\ -2 \end{bmatrix} } \begin{bmatrix} 7\\ 7\\ 5\\ 5\\ -2 \end{bmatrix} = \begin{pmatrix}\frac{1}{4}\\ \frac{1}{4}\\ -\frac{1}{4}\\ -\frac{1}{4}\\ \frac{1}{2}\end{pmatrix}$$

Multiply $$\textbf {y}_3$$ by 4 to replace it by a vector without fractions.
$$\textbf {y'}_3 = 4 \textbf {y}_3 = \begin{bmatrix} 1\\ 1 \\ -1 \\ -1\\ 2 \end{bmatrix}$$

Use $$\textbf {y'}_3$$ instead of $$\textbf {y}_3$$ in the formulas

$$\textbf {y}_4 = \textbf {v}_4 - \dfrac{\textbf {v}_4 \cdot \textbf {y}_1}{\textbf {y}_1 \cdot \textbf {y}_1} \textbf {y}_1 - \dfrac{\textbf {v}_4 \cdot \textbf {y'}_2}{\textbf {y'}_2 \cdot \textbf {y'}_2} \textbf {y'}_2 - \dfrac{\textbf {v}_4\cdot \textbf {y'}_3}{\textbf {y'}_3 \cdot \textbf {y'}_3} \textbf {y'}_3$$

Substitute
$$\textbf {y}_4 = \begin{bmatrix} 0 \\ 0 \\ 0\\ 1\\ 1 \end{bmatrix} - \dfrac{\begin{bmatrix} 0 \\ 0 \\ 0\\ 1\\ 1 \end{bmatrix} \cdot \begin{bmatrix} -1\\ -1 \\ 2 \\ 2\\ 3 \end{bmatrix}}{\begin{bmatrix} -1\\ -1 \\ 2 \\ 2\\ 3 \end{bmatrix} \cdot \begin{bmatrix} -1\\ -1 \\ 2 \\ 2\\ 3 \end{bmatrix}} \begin{bmatrix} -1\\ -1 \\ 2 \\ 2\\ 3 \end{bmatrix}$$

- $$\dfrac{\begin{bmatrix} 0 \\ 0 \\ 0\\ 1\\ 1 \end{bmatrix} \cdot \begin{bmatrix} 7\\ 7\\ 5\\ 5\\ -2 \end{bmatrix}}{\begin{bmatrix} 7\\ 7\\ 5\\ 5\\ -2 \end{bmatrix} \cdot \begin{bmatrix} 7\\ 7\\ 5\\ 5\\ -2 \end{bmatrix}} \begin{bmatrix} 7\\ 7\\ 5\\ 5\\ -2 \end{bmatrix} \\ \quad \quad - \dfrac{\begin{bmatrix} 0 \\ 0 \\ 0\\ 1\\ 1 \end{bmatrix} \cdot \begin{bmatrix} 1\\ 1 \\ -1 \\ -1\\ 2 \end{bmatrix}}{\begin{bmatrix} 1\\ 1 \\ -1 \\ -1\\ 2 \end{bmatrix} \cdot \begin{bmatrix} 1\\ 1 \\ -1 \\ -1\\ 2 \end{bmatrix}} \begin{bmatrix} 1\\ 1 \\ -1 \\ -1\\ 2 \end{bmatrix} = \begin{pmatrix}0\\ 0\\ -\frac{1}{2}\\ \frac{1}{2}\\ 0\end{pmatrix}$$
Multiply $$\textbf {y}_4$$ by 2 to obtain
$$\textbf {y'}_4 = 2 \textbf {y}_4 = \begin{bmatrix} 0\\ 0\\ -1 \\ 1\\ 0 \end{bmatrix}$$
Hence an
orthogonal basis for the subset $$V$$ may be written as
$$Y = \left \{ \textbf {y}_1 , \textbf {y'}_2 , \textbf {y'}_3 , \textbf {y'}_4 \right \} = \left \{ \begin{bmatrix} -1\\ -1 \\ 2 \\ 2\\ 3 \end{bmatrix} , \begin{bmatrix} 7\\ 7\\ 5\\ 5\\ -2 \end{bmatrix} , \begin{bmatrix} 1\\ 1 \\ -1 \\ -1\\ 2 \end{bmatrix} , \begin{bmatrix} 0\\ 0\\ -1 \\ 1\\ 0 \end{bmatrix} \right \}$$
The
orthonormal basis $$Y_O$$ is obtained by dividing each vector in the basis $$Y$$ by its norm.
$$Y_O = \left \{ \dfrac{1}{\sqrt {19}} \begin{bmatrix} -1\\ -1 \\ 2 \\ 2\\ 3 \end{bmatrix} , \dfrac{1}{2\sqrt{38}} \begin{bmatrix} 7\\ 7\\ 5\\ 5\\ -2 \end{bmatrix} , \dfrac{1}{2\sqrt{2}} \begin{bmatrix} 1\\ 1 \\ -1 \\ -1\\ 2 \end{bmatrix} , \dfrac{1}{\sqrt{2}} \begin{bmatrix} 0\\ 0\\ -1 \\ 1\\ 0 \end{bmatrix} \right \}$$

To check that the basis $$Y$$ obtained span that same subspace as the given basis $$V$$, we row reduce the matrix
$$\begin{bmatrix} V | Y \end{bmatrix} = \begin{bmatrix}-1&1&0&0&-1&7&1&0\\ \:\:\:-1&1&0&0&-1&7&1&0\\ \:\:\:2&1&-1&0&2&5&-1&-1\\ \:\:\:2&1&-1&1&2&5&-1&1\\ \:\:\:3&0&0&1&3&-1&2&0\end{bmatrix}$$
to obtain
$$\begin{bmatrix}1&0&0&0&1&-\frac{1}{3}&\frac{2}{3}&-\frac{2}{3}\\ 0&1&0&0&0&\frac{20}{3}&\frac{5}{3}&-\frac{2}{3}\\ 0&0&1&0&0&1&4&-1\\ 0&0&0&1&0&0&0&2\\ 0&0&0&0&0&0&0&0\end{bmatrix}$$
and conlude from the above that
$$y_1 = v_1$$
$$y_2 = - \dfrac {1}{3} v_1 + \dfrac{20}{3} v_2 + v_3$$
$$y_3 = \dfrac {2}{3} v_1 + \dfrac{5}{3} v_2 + 4 v_3$$
$$y_4 = - \dfrac {2}{3} v_1 - \dfrac{2}{3} v_2 - v_3 + 2 v_4$$
The above results show that span $$\{ \textbf {v}_1, \textbf{v}_2,\textbf{v}_3 ,\textbf{v}_4 \}$$ = span $$\{ \textbf {y}_1, \textbf{y}_2, \textbf{y}_3 , \textbf{y}_3 \}$$
Using the inner product, it can easily be shown that $$Y$$ is an orthogonal basis.

## More References and links

1. Vector Spaces - Questions with Solutions
2. Linear Algebra and its Applications - 5 th Edition - David C. Lay , Steven R. Lay , Judi J. McDonald
3. Elementary Linear Algebra - 7 th Edition - Howard Anton and Chris Rorres
4. Introduction to Linear Algebra - Fifth Edition (2016) - Gilbert Strang
5. Linear Algebra Done Right - third edition, 2015 - Sheldon Axler
6. Linear Algebra with Applications - 2012 - Gareth Williams

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