(a) Let I3 denote the 3 x 3 identity matrix. What should be the shape of in order for I3 to be computable? (b) I3 contains 3 column vectors e₁,e2, e3. Let x be a 3-vector [x1 2. Rewrite x in the form of a linear combination of e₁, €2, €3. (c) Part b tells us that I3 is a basis for R³ because any 3-vector can be written as a linear combination of the column vectors of I3. 1 However, I is not the only basis of R3. Prove that B = 0 0 linearly independent. 0 1 1 1 1 0 is a basis for R3 by showing that the columns of B are
(a) Let I3 denote the 3 x 3 identity matrix. What should be the shape of in order for I3 to be computable? (b) I3 contains 3 column vectors e₁,e2, e3. Let x be a 3-vector [x1 2. Rewrite x in the form of a linear combination of e₁, €2, €3. (c) Part b tells us that I3 is a basis for R³ because any 3-vector can be written as a linear combination of the column vectors of I3. 1 However, I is not the only basis of R3. Prove that B = 0 0 linearly independent. 0 1 1 1 1 0 is a basis for R3 by showing that the columns of B are
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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Question
Represent vectors using basis
More details.
![**(a)** Let \( I_3 \) denote the \( 3 \times 3 \) identity matrix. What should be the shape of \( x \) in order for \( I_3 x \) to be computable?
---
**(b)** \( I_3 \) contains 3 column vectors \( e_1, e_2, e_3 \). Let \( x \) be a 3-vector
\[
\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}
\]
. Rewrite \( x \) in the form of a linear combination of \( e_1, e_2, e_3 \).
---
**(c)** Part b tells us that \( I_3 \) is a basis for \( \mathbb{R}^3 \) because any 3-vector can be written as a linear combination of the column vectors of \( I_3 \).
However, \( I_3 \) is not the only basis of \( \mathbb{R}^3 \). Prove that
\[
B = \begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 1 & 0
\end{bmatrix}
\]
is a basis for \( \mathbb{R}^3 \) by showing that the columns of \( B \) are linearly independent.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa98117fc-4537-4efb-9016-9993926a8208%2Fdf2ef4a7-9b90-435d-b7fa-0b433977ea24%2F9ttfkkl_processed.png&w=3840&q=75)
Transcribed Image Text:**(a)** Let \( I_3 \) denote the \( 3 \times 3 \) identity matrix. What should be the shape of \( x \) in order for \( I_3 x \) to be computable?
---
**(b)** \( I_3 \) contains 3 column vectors \( e_1, e_2, e_3 \). Let \( x \) be a 3-vector
\[
\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}
\]
. Rewrite \( x \) in the form of a linear combination of \( e_1, e_2, e_3 \).
---
**(c)** Part b tells us that \( I_3 \) is a basis for \( \mathbb{R}^3 \) because any 3-vector can be written as a linear combination of the column vectors of \( I_3 \).
However, \( I_3 \) is not the only basis of \( \mathbb{R}^3 \). Prove that
\[
B = \begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 1 & 0
\end{bmatrix}
\]
is a basis for \( \mathbb{R}^3 \) by showing that the columns of \( B \) are linearly independent.
![(d) Normally, the 3-vector we write down uses basis \( I_3 \). For example,
\[ b = \begin{bmatrix} 2 \\ 3 \\ 6 \end{bmatrix} \]
\[ = 2 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + 3 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + 6 \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \]
uses basis \( I_3 \). Now express vector \( b \) using basis \( B \). In other words, you are trying to find scalars \( z_1, z_2, z_3 \) such that
\[ \begin{bmatrix} 2 \\ 3 \\ 6 \end{bmatrix} = z_1 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + z_2 \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + z_3 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \]
Finally,
\[ z = \begin{bmatrix} z_1 \\ z_2 \\ z_3 \end{bmatrix} \]
is how you represent \( b \) using basis \( B \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa98117fc-4537-4efb-9016-9993926a8208%2Fdf2ef4a7-9b90-435d-b7fa-0b433977ea24%2Flmwd4e8_processed.png&w=3840&q=75)
Transcribed Image Text:(d) Normally, the 3-vector we write down uses basis \( I_3 \). For example,
\[ b = \begin{bmatrix} 2 \\ 3 \\ 6 \end{bmatrix} \]
\[ = 2 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + 3 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + 6 \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \]
uses basis \( I_3 \). Now express vector \( b \) using basis \( B \). In other words, you are trying to find scalars \( z_1, z_2, z_3 \) such that
\[ \begin{bmatrix} 2 \\ 3 \\ 6 \end{bmatrix} = z_1 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + z_2 \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + z_3 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \]
Finally,
\[ z = \begin{bmatrix} z_1 \\ z_2 \\ z_3 \end{bmatrix} \]
is how you represent \( b \) using basis \( B \).
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Follow-up Question
Represent vectors using basis
More details. Just need (d)
![(d) Normally, the 3-vector we write down uses basis \( I_3 \). For example,
\[ b = \begin{bmatrix} 2 \\ 3 \\ 6 \end{bmatrix} \]
\[ = 2 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + 3 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + 6 \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \]
uses basis \( I_3 \). Now express vector \( b \) using basis \( B \). In other words, you are trying to find scalars \( z_1, z_2, z_3 \) such that
\[ \begin{bmatrix} 2 \\ 3 \\ 6 \end{bmatrix} = z_1 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + z_2 \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + z_3 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \]
Finally,
\[ z = \begin{bmatrix} z_1 \\ z_2 \\ z_3 \end{bmatrix} \]
is how you represent \( b \) using basis \( B \).](https://content.bartleby.com/qna-images/question/a98117fc-4537-4efb-9016-9993926a8208/67594587-452d-4ce5-9ae4-24342b573257/8uz6jf_processed.png)
Transcribed Image Text:(d) Normally, the 3-vector we write down uses basis \( I_3 \). For example,
\[ b = \begin{bmatrix} 2 \\ 3 \\ 6 \end{bmatrix} \]
\[ = 2 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + 3 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + 6 \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \]
uses basis \( I_3 \). Now express vector \( b \) using basis \( B \). In other words, you are trying to find scalars \( z_1, z_2, z_3 \) such that
\[ \begin{bmatrix} 2 \\ 3 \\ 6 \end{bmatrix} = z_1 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + z_2 \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + z_3 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \]
Finally,
\[ z = \begin{bmatrix} z_1 \\ z_2 \\ z_3 \end{bmatrix} \]
is how you represent \( b \) using basis \( B \).
![(a) Let \( I_3 \) denote the \( 3 \times 3 \) identity matrix. What should be the shape of \( \mathbf{x} \) in order for \( I_3 \mathbf{x} \) to be computable?
---
(b) \( I_3 \) contains 3 column vectors \( \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3 \). Let \( \mathbf{x} \) be a 3-vector
\[
\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}
\]
Rewrite \( \mathbf{x} \) in the form of a linear combination of \( \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3 \).
---
(c) Part (b) tells us that \( I_3 \) is a basis for \( \mathbb{R}^3 \) because any 3-vector can be written as a linear combination of the column vectors of \( I_3 \).
However, \( I_3 \) is not the only basis of \( \mathbb{R}^3 \). Prove that \( B = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} \) is a basis for \( \mathbb{R}^3 \) by showing that the columns of \( B \) are linearly independent.](https://content.bartleby.com/qna-images/question/a98117fc-4537-4efb-9016-9993926a8208/67594587-452d-4ce5-9ae4-24342b573257/8eddknk_processed.png)
Transcribed Image Text:(a) Let \( I_3 \) denote the \( 3 \times 3 \) identity matrix. What should be the shape of \( \mathbf{x} \) in order for \( I_3 \mathbf{x} \) to be computable?
---
(b) \( I_3 \) contains 3 column vectors \( \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3 \). Let \( \mathbf{x} \) be a 3-vector
\[
\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}
\]
Rewrite \( \mathbf{x} \) in the form of a linear combination of \( \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3 \).
---
(c) Part (b) tells us that \( I_3 \) is a basis for \( \mathbb{R}^3 \) because any 3-vector can be written as a linear combination of the column vectors of \( I_3 \).
However, \( I_3 \) is not the only basis of \( \mathbb{R}^3 \). Prove that \( B = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} \) is a basis for \( \mathbb{R}^3 \) by showing that the columns of \( B \) are linearly independent.
Solution
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