Vector Spaces - The Space F^n - Cooperstein Exercise 12, page 10

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SUMMARY

The discussion centers on Exercise 12 from Bruce Cooperstein's book, "Advanced Linear Algebra," specifically regarding the space $$\mathbb{F}^n$$. The user, Peter, attempts to solve the equation $$\begin{pmatrix} 2v_1 \\ 2v_2 \end{pmatrix} + \begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$$ (mod 5) and arrives at the solution $$\underline{v} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$$. However, the book's answer states $$\underline{v} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$$. Other participants in the discussion suggest that the discrepancy lies with Cooperstein's solution rather than Peter's calculations.

PREREQUISITES
  • Understanding of modular arithmetic, specifically mod 5 operations.
  • Familiarity with vector spaces, particularly $$\mathbb{F}^n$$.
  • Basic knowledge of linear algebra concepts and equations.
  • Experience with solving linear equations in vector form.
NEXT STEPS
  • Review modular arithmetic principles, focusing on operations mod 5.
  • Study vector space properties and operations in $$\mathbb{F}^n$$.
  • Examine common errors in solving linear equations in modular contexts.
  • Explore additional exercises in Bruce Cooperstein's "Advanced Linear Algebra" for further practice.
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Students of linear algebra, educators teaching modular arithmetic, and anyone seeking to clarify concepts related to vector spaces and their applications in mathematical problems.

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I am reading Bruce Cooperstein's book: Advanced Linear Algebra ... ...

I am focused on Section 1.2 The Space $$\mathbb{F}^n$$ ...

I need help with Exercise 12 ... since I do not get the same answer as the author ...

Exercise 12 reads as follows:

View attachment 5103My attempt at a solution to this apparently simple problem is as follows:
Let $$\underline{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$$ ... ... We need to find $$v_1$$ and $$v_2$$ such that:

$$\begin{pmatrix} 2v_1 \\ 2v_2 \end{pmatrix} + \begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$$ (mod 5) ... ... ... (1)Thus, given (1), we must have, for $$v_1$$ :

$$2 v_1 +3 = 1$$ (mod 5)

so we must have

$$2 v_1 = 1 - 3 = -2$$ (mod 5)

and so

$$v_1 = -1 \equiv 4$$ (mod 5)

----------------------------------------------

For $$v_2$$, given (1)we must have

$$2 v_2 + 4 = 3$$ (mod 5)

So then we must have

$$2 v_2 = 3 - 4 = -1 \equiv 4$$ (mod 5)

and so

$$v_2 = 2$$ Thus ...

$$\underline{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$$BUT ... in the "Hints to Selected Problems, Cooperstein gives the answer as$$\underline{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$$
So, my problem is ... why is there a discrepancy ... where is my error ...?

Hope someone can help ...

Peter
 
Physics news on Phys.org
It looks as though the error is Cooperstein's, not yours.
 

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