-25 +-0-0-8} U₁ 3 U₂ 25 U3 25 find an orthogonal basis under the Euclidean inner product. Let Orthogonal basis: V₁ = a = Ex: 1.23 -21 {~--~-~-0} 25 b = Ex: 1.23 be a basis for R³. Use the Gram-Schmidt process to c = Ex: 1.23
-25 +-0-0-8} U₁ 3 U₂ 25 U3 25 find an orthogonal basis under the Euclidean inner product. Let Orthogonal basis: V₁ = a = Ex: 1.23 -21 {~--~-~-0} 25 b = Ex: 1.23 be a basis for R³. Use the Gram-Schmidt process to c = Ex: 1.23
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Gram-Schmidt Process Example
#### Given Basis Vectors:
Let
\[
\mathbf{u}_1 = \begin{pmatrix} 4 \\ 3 \\ 0 \end{pmatrix}, \quad \mathbf{u}_2 = \begin{pmatrix} -25 \\ 25 \\ 25 \end{pmatrix}, \quad \mathbf{u}_3 = \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix}
\]
be a basis for \(\mathbb{R}^3\). Use the Gram-Schmidt process to find an orthogonal basis under the Euclidean inner product.
#### Orthogonal Basis:
\[
\mathbf{v}_1 = \begin{pmatrix} 4 \\ 3 \\ 0 \end{pmatrix}, \quad \mathbf{v}_2 = \begin{pmatrix} -21 \\ 28 \\ 25 \end{pmatrix}, \quad \mathbf{v}_3 = \begin{pmatrix} a \\ b \\ c \end{pmatrix}
\]
#### Values for Constants:
\[
a = \text{Ex: 1.23}, \quad b = \text{Ex: 1.23}, \quad c = \text{Ex: 1.23}
\]
##### Explanation:
The text provides an example of applying the Gram-Schmidt process to a set of vectors to obtain an orthogonal basis. The vectors \(\mathbf{u}_1\), \(\mathbf{u}_2\), and \(\mathbf{u}_3\) are given as a basis for \(\mathbb{R}^3\). The Gram-Schmidt process starts with \(\mathbf{v}_1\) identical to \(\mathbf{u}_1\), derives \(\mathbf{v}_2\) from \(\mathbf{u}_2\) orthogonalized to \(\mathbf{v}_1\), and then \(\mathbf{v}_3\) is orthogonalized from both \(\mathbf{v}_1\) and \(\mathbf{v}_2\). The constants \(a\), \(b\), and \(c\) would complete \(\mathbf{v}_3\), but they are placeholders in this example.
This demonstration is aimed at helping students understand how to use the Gram-Schmidt process to convert](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf8aa82a-31a1-4382-a30e-3ad6a3174f84%2Ffa9f102c-ad20-4596-9405-f91441cde865%2F4lr30sw_processed.png&w=3840&q=75)
Transcribed Image Text:### Gram-Schmidt Process Example
#### Given Basis Vectors:
Let
\[
\mathbf{u}_1 = \begin{pmatrix} 4 \\ 3 \\ 0 \end{pmatrix}, \quad \mathbf{u}_2 = \begin{pmatrix} -25 \\ 25 \\ 25 \end{pmatrix}, \quad \mathbf{u}_3 = \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix}
\]
be a basis for \(\mathbb{R}^3\). Use the Gram-Schmidt process to find an orthogonal basis under the Euclidean inner product.
#### Orthogonal Basis:
\[
\mathbf{v}_1 = \begin{pmatrix} 4 \\ 3 \\ 0 \end{pmatrix}, \quad \mathbf{v}_2 = \begin{pmatrix} -21 \\ 28 \\ 25 \end{pmatrix}, \quad \mathbf{v}_3 = \begin{pmatrix} a \\ b \\ c \end{pmatrix}
\]
#### Values for Constants:
\[
a = \text{Ex: 1.23}, \quad b = \text{Ex: 1.23}, \quad c = \text{Ex: 1.23}
\]
##### Explanation:
The text provides an example of applying the Gram-Schmidt process to a set of vectors to obtain an orthogonal basis. The vectors \(\mathbf{u}_1\), \(\mathbf{u}_2\), and \(\mathbf{u}_3\) are given as a basis for \(\mathbb{R}^3\). The Gram-Schmidt process starts with \(\mathbf{v}_1\) identical to \(\mathbf{u}_1\), derives \(\mathbf{v}_2\) from \(\mathbf{u}_2\) orthogonalized to \(\mathbf{v}_1\), and then \(\mathbf{v}_3\) is orthogonalized from both \(\mathbf{v}_1\) and \(\mathbf{v}_2\). The constants \(a\), \(b\), and \(c\) would complete \(\mathbf{v}_3\), but they are placeholders in this example.
This demonstration is aimed at helping students understand how to use the Gram-Schmidt process to convert
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