Solving Linear Transformations in R2: Step by Step Guide

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SUMMARY

This discussion provides a step-by-step method for solving linear transformations in the vector space R2, specifically focusing on the transformation phi defined by phi((1,0)) = (4,5) and phi((0,1)) = (9,11). The transformation of a general vector (x,y) is expressed as phi(x,y) = (4x + 9y, 5x + 11y). Additionally, the discussion details how to determine the matrix M_B^B(phi) with respect to the ordered basis B formed by the vectors w1 = (3,5) and w2 = (10,17), leading to the calculation of coefficients α1, α2, β1, and β2 through a system of equations.

PREREQUISITES
  • Understanding of linear transformations in vector spaces
  • Familiarity with matrix representation of linear maps
  • Knowledge of ordered bases in R2
  • Ability to solve systems of linear equations
NEXT STEPS
  • Study the properties of linear transformations in R2
  • Learn how to compute the matrix representation of linear maps
  • Explore the concept of change of basis in linear algebra
  • Practice solving systems of linear equations using matrix methods
USEFUL FOR

Students preparing for exams in linear algebra, educators teaching vector spaces, and anyone interested in mastering linear transformations and their matrix representations.

Fernando Revilla
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I quote a question from Yahoo! Answers

Could anyone give me a step by step method on how to solve these type of questions please? Very confused and need to know for my exams.

Let V denote the real vector space R2 and phi: V -> V be a real linear transformation such that phi((1,0)) = (4,5) and phi((0,1)) = (9,11). Express the imagine phi((x,y)) of (x,y) in terms of x and y.

Assume that w1 = (3,5) and w2 = (10,17) form an ordered basis B for V. Working from the definition determine the matrix M(subscript and superscript B) (phi) of phi with respect to the basis B.

I have given a link to the topic there so the OP can see my response.
 
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Using that $\phi$ is a linear map:
$$\phi(x,y)=\phi[x(1,0)+y(0,1)]=x\phi (1,0)+y\phi(0,1)\\=x(4,5)+y(9,11)=(4x+9y,5x+11y)$$
On the other hand:
$$\phi(w_1)=\phi (3,5)=(4\cdot 3+9\cdot 5,\;5\cdot 3+11\cdot 5)=(57,70)$$
Now, write $\phi(w_1)=(57,70)=\alpha_1w_1+\alpha_2 w_2$ and you'll easily find $\alpha_1$ and $\alpha_2$ by means of a simple system. In a similar way, you'll get $\phi(w_2)=\beta_1w_1+\beta_2 w_2$. Then,
$$M_B^B= \begin{bmatrix}{\alpha_1}&{\beta_1}\\{\alpha_2}&{ \beta_2}\end{bmatrix}=\ldots$$
 
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