**Problem 5: Consider Two Basis of $ \mathbb{R}^2 $.**Given:- Basis $ B = \left\{ \begin{pmatrix} 7 \\ -1 \end{pmatrix}, \begin{pmatrix} -3 \\ -1 \end{pmatrix} \right\} $- Basis $ C = \left\{ \begin{pmatrix} 3 \\ -5 \end{pmatrix}, \begin{pmatrix} -1 \\ 2 \end{pmatrix} \right\} $Question:- If a vector $ \vec{v} \in \mathbb{R}^2 $ has coordinates $ \begin{pmatrix} 2 \\ 1 \end{pmatrix} $ with respect to basis $ B $, what are its coordinates with respect to basis $ C $?

The given 2 bases of ℝ2 are as follows. B=75, -3-1 and C=3-5, -12 Also, the coordinates of the…

Answered: 2. 5. Consider twO basis of R: B = %3D…

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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**Problem 5: Consider Two Basis of $ \mathbb{R}^2 $.**

Given:
- Basis $ B = \left\{ \begin{pmatrix} 7 \\ -1 \end{pmatrix}, \begin{pmatrix} -3 \\ -1 \end{pmatrix} \right\} $

- Basis $ C = \left\{ \begin{pmatrix} 3 \\ -5 \end{pmatrix}, \begin{pmatrix} -1 \\ 2 \end{pmatrix} \right\} $

Question:
- If a vector $ \vec{v} \in \mathbb{R}^2 $ has coordinates $ \begin{pmatrix} 2 \\ 1 \end{pmatrix} $ with respect to basis $ B $, what are its coordinates with respect to basis $ C $?

Expert Solution

Step 1

The given 2 bases of $ℝ^{2}$ are as follows.

$B = \{(\begin{matrix} 7 \\ 5 \end{matrix}), (\begin{matrix} - 3 \\ - 1 \end{matrix})\}$ and $C = \{(\begin{matrix} 3 \\ - 5 \end{matrix}), (\begin{matrix} - 1 \\ 2 \end{matrix})\}$

Also, the coordinates of the vector v with respect to the basis B is given by $(\begin{matrix} 2 \\ 1 \end{matrix})$ .

This implies that the vector $v = 2 (\begin{matrix} 7 \\ 5 \end{matrix}) + 1 (\begin{matrix} - 3 \\ - 1 \end{matrix})$ .

Hence, we have

$\begin{array}{rcl} v & = & 2 (\begin{matrix} 7 \\ 5 \end{matrix}) + 1 (\begin{matrix} - 3 \\ - 1 \end{matrix}) \\ = & (\begin{matrix} 14 \\ 10 \end{matrix}) + (\begin{matrix} - 3 \\ - 1 \end{matrix}) \\ = & (\begin{matrix} 11 \\ 9 \end{matrix}) \end{array}$

Step 2

Let $(\begin{matrix} a \\ b \end{matrix})$ be the coordinates of the vector v with respect to the basis C.

Then,

$\begin{array}{rcl} v & = & a (\begin{matrix} 3 \\ - 5 \end{matrix}) + b (\begin{matrix} - 1 \\ 2 \end{matrix}) \\ = & (\begin{matrix} 3 a \\ - 5 a \end{matrix}) + (\begin{matrix} - b \\ 2 b \end{matrix}) \\ = & (\begin{matrix} 3 a - b \\ - 5 a + 2 b \end{matrix}) \end{array}$

Substituting the vector $\begin{array}{rcl} v & = & (\begin{matrix} 11 \\ 9 \end{matrix}) \end{array}$ in $\begin{array}{rcl} v & = & (\begin{matrix} 3 a - b \\ - 5 a + 2 b \end{matrix}) \end{array}$ , we get $\begin{array}{rcl} (\begin{matrix} 11 \\ 9 \end{matrix}) & = & (\begin{matrix} 3 a - b \\ - 5 a + 2 b \end{matrix}) \end{array}$ .

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