Chapter 1: Q31E (page 1)

Perform the Gram-Schmidt process on the following basis of $R^{3}$ ${\vec{v}}_{1} = [\begin{matrix} a \\ 0 \\ 0 \end{matrix}]$ , ${\vec{v}}_{2} = [\begin{matrix} b \\ c \\ 0 \end{matrix}]$ and ${\vec{v}}_{3} = [\begin{matrix} d \\ e \\ f \end{matrix}]$ Illustrate the work with a sketch.

Short Answer

Expert verified

The solution is the orthonormal vectors are $\{{\vec{e}}_{1}, {\vec{e}}_{2}, {\vec{e}}_{3}\}$ .

Step by step solution

`Explanation of the solution

Consider the vectors as follows.

$\{{\vec{v}}_{1}, {\vec{v}}_{2}, {\vec{v}}_{3}\}$

To perform the Gram-Schmidt process to obtain orthonormal vectors.

According to the Gram-Schmidt process as follows.

localid="1660119238407" ${\vec{u}}_{1} = \frac{{\vec{v}}_{1}}{||\vec{v}||} {\vec{u}}_{2} = \frac{{\vec{v}}_{2} - (\vec{u} \cdot {\vec{v}}_{2}) {\vec{u}}_{1}}{||{\vec{v}}_{2} - (\vec{u} \cdot {\vec{v}}_{2}) {\vec{u}}_{1}||} {\vec{u}}_{3} = \frac{{\vec{v}}_{3} - ({\vec{u}}_{1} \cdot {\vec{v}}_{3}) {\vec{u}}_{1} - ({\vec{u}}_{2} \cdot {\vec{v}}_{3}) {\vec{u}}_{2}}{||{\vec{v}}_{2} - (\vec{u} \cdot {\vec{v}}_{2}) {\vec{u}}_{1} - ({\vec{u}}_{2} \cdot {\vec{v}}_{3}) {\vec{u}}_{2}||}$

Since, $||{\vec{v}}_{1}|| = |a| = a$

So, $||{\vec{u}}_{1}|| = [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}] = {\vec{e}}_{1}$ .

Now, in order to computelocalid="1660119335550" ${\vec{u}}_{2}$ as follows.

${\vec{u}}_{1} \cdot {\vec{v}}_{2} = b ({\vec{u}}_{1} \cdot {\vec{v}}_{2}) {\vec{u}}_{1} = \frac{1}{2} [\begin{matrix} b \\ 0 \\ 0 \end{matrix}] {\vec{v}}_{2} - ({\vec{u}}_{1} \cdot {\vec{v}}_{2}) {\vec{u}}_{1} = [\begin{matrix} b \\ c \\ 0 \end{matrix}] - [\begin{matrix} b \\ 0 \\ 0 \end{matrix}] {\vec{v}}_{2} - ({\vec{u}}_{1} \cdot {\vec{v}}_{2}) {\vec{u}}_{1} = [\begin{matrix} 0 \\ c \\ 0 \end{matrix}]$

Simplify further as follows.

$||{\vec{v}}_{2} - ({\vec{u}}_{1} \cdot {\vec{v}}_{2}) {\vec{u}}_{1}|| = |c| = c {\vec{u}}_{2} = {\vec{e}}_{2}$

In order to find ${\vec{u}}_{3}$ need to compute the following.

${\vec{u}}_{1} \cdot {\vec{v}}_{3} = d ({\vec{u}}_{1} \cdot {\vec{v}}_{3}) {\vec{u}}_{1} = \frac{1}{2} [\begin{matrix} d \\ 0 \\ 0 \end{matrix}] {\vec{u}}_{2} \cdot {\vec{v}}_{3} = e ({\vec{u}}_{2} {\vec{v}}_{3}) {\vec{u}}_{2} = [\begin{matrix} 0 \\ e \\ 0 \end{matrix}] {\vec{v}}_{3} - ({\vec{u}}_{1} \cdot {\vec{v}}_{3}) {\vec{u}}_{1} - ({\vec{u}}_{2} \cdot {\vec{v}}_{3}) {\vec{u}}_{2} = [\begin{matrix} d \\ e \\ f \end{matrix}] - [\begin{matrix} d \\ 0 \\ 0 \end{matrix}] - [\begin{matrix} 0 \\ e \\ 0 \end{matrix}] {\vec{v}}_{3} - ({\vec{u}}_{1} \cdot {\vec{v}}_{3}) {\vec{u}}_{1} - ({\vec{u}}_{2} \cdot {\vec{v}}_{3}) {\vec{u}}_{2} = [\begin{matrix} 0 \\ 0 \\ f \end{matrix}] ||{\vec{v}}_{3} - ({\vec{u}}_{1} \cdot {\vec{v}}_{3}) {\vec{u}}_{1} - ({\vec{u}}_{2} \cdot {\vec{v}}_{3}) {\vec{u}}_{2}|| = |f| = f {\vec{u}}_{3} = \frac{1}{2} [\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}] = {\vec{e}}_{3}$

Simplify further as follows.

${\vec{v}}_{3} - ({\vec{u}}_{1} \cdot {\vec{v}}_{3}) {\vec{u}}_{1} - ({\vec{u}}_{2} \cdot {\vec{v}}_{3}) {\vec{u}}_{2} = [\begin{matrix} d \\ e \\ f \end{matrix}] - [\begin{matrix} d \\ 0 \\ 0 \end{matrix}] - [\begin{matrix} 0 \\ e \\ 0 \end{matrix}] {\vec{v}}_{3} - ({\vec{u}}_{1} \cdot {\vec{v}}_{3}) {\vec{u}}_{1} - ({\vec{u}}_{2} \cdot {\vec{v}}_{3}) {\vec{u}}_{2} = [\begin{matrix} 0 \\ 0 \\ f \end{matrix}] ||{\vec{v}}_{3} - ({\vec{u}}_{1} \cdot {\vec{v}}_{3}) {\vec{u}}_{1} - ({\vec{u}}_{2} \cdot {\vec{v}}_{3}) {\vec{u}}_{2}|| = |f| = f {\vec{u}}_{3} = \frac{1}{2} [\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}] = {\vec{e}}_{3}$

Therefore, the orthonormal vectors are $\{{\vec{e}}_{1}, {\vec{e}}_{2}, {\vec{e}}_{3}\}$ .

Graphical representation of the solution

The orthonormal vectors $\{{\vec{e}}_{1}, {\vec{e}}_{2}, {\vec{e}}_{3}\}$ are sketched in the graph as follows.

Thus, the orthonormal vectors $\{{\vec{e}}_{1}, {\vec{e}}_{2}, {\vec{e}}_{3}\}$ are sketched in the graph.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Perform the Gram-Schmidt process on the following basis of $R^{3}$ ${\vec{v}}_{1} = [\begin{matrix} a \\ 0 \\ 0 \end{matrix}]$ , ${\vec{v}}_{2} = [\begin{matrix} b \\ c \\ 0 \end{matrix}]$ and ${\vec{v}}_{3} = [\begin{matrix} d \\ e \\ f \end{matrix}]$ Illustrate the work with a sketch.

Short Answer

Step by step solution

`Explanation of the solution

Graphical representation of the solution

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Statistics

Theoretical and Mathematical Physics

Calculus

Decision Maths

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Company

Product

Help

Perform the Gram-Schmidt process on the following basis of R3v→1=[a00],v→2=[bc0] andv→3=[def] Illustrate the work with a sketch.

Short Answer

Step by step solution

`Explanation of the solution

Graphical representation of the solution

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Statistics

Theoretical and Mathematical Physics

Calculus

Decision Maths

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Perform the Gram-Schmidt process on the following basis of $R^{3}$ ${\vec{v}}_{1} = [\begin{matrix} a \\ 0 \\ 0 \end{matrix}]$ , ${\vec{v}}_{2} = [\begin{matrix} b \\ c \\ 0 \end{matrix}]$ and ${\vec{v}}_{3} = [\begin{matrix} d \\ e \\ f \end{matrix}]$ Illustrate the work with a sketch.