Linear Algebra: Matrix Transformations

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SUMMARY

The discussion focuses on matrix transformations defined by the function f: R3 → R2, specifically using the matrix A = \begin{pmatrix} 1 & 2 & 0\\ 0 & 1 & -1\\ \end{pmatrix} to find different vectors u and v such that f(u) = f(v) = w, where w = \begin{pmatrix} 0\\ -1\\ \end{pmatrix}. The key solution involves solving the linear system of equations derived from the transformation, which reveals a free variable that allows for multiple solutions. The discussion also clarifies the notation of R2 and R3 in the context of vector spaces.

PREREQUISITES
  • Understanding of matrix multiplication and transformations
  • Familiarity with linear systems of equations
  • Knowledge of vector spaces R2 and R3
  • Basic concepts of free variables in linear algebra
NEXT STEPS
  • Study the properties of linear transformations in linear algebra
  • Learn to solve linear systems using Gaussian elimination
  • Explore the concept of free variables and their implications in vector spaces
  • Investigate matrix rank and its relationship to the solutions of linear equations
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Students of linear algebra, educators teaching matrix transformations, and anyone seeking to deepen their understanding of vector spaces and linear systems.

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Homework Statement


Some matrix transformations [itex]f[/itex] have the property that [itex]f(u) = f(v), when u ≠v[/itex]. That is, the images of different vectors can be the same. For each of the following matrix transformations [itex]f : R^{2} → R^{2}[/itex] defined by [itex]f(u) = Au[/itex], find two different vectors [itex]u[/itex] and [itex]v[/itex] such that [itex]f(u)=f(v)=w[/itex] for the given vector [itex]w[/itex].

A = [tex] \begin{pmatrix}<br /> 1 & 2 & 0\\<br /> 0 & 1 & -1\\<br /> \end{pmatrix}[/tex]
w=
[tex] \begin{pmatrix}<br /> 0\\<br /> -1\\<br /> \end{pmatrix}[/tex]

Homework Equations



My professor noted that there was a typo in the book, and that instead of [itex]f : R^{2} → R^{2}[/itex], it should be [itex]f : R^{3} → R^{2}[/itex].

The Attempt at a Solution



My professor has been on jury duty for the past week, and our sub just assigns us homework without much instruction or guidance. We haven't been properly introduced to the notation [itex]R[/itex], either.

But from inferring from the problem, is it safe to assume that [itex]w = Au[/itex]?
Also, would I have to use something along the lines of
u = [tex] \begin{pmatrix}<br /> x\\<br /> y\\<br /> z\\<br /> \end{pmatrix}[/tex] ?

Detailed instruction would be much appreciated, as I am anxious to grasp the subject matter.
 
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yes w = Au means that the (column) vector w is equal to the matrix product of the matrix A and the (column) vector u.

so if u = (x,y,z)T, then Au = (x+2y,y-z)T

so you need to find 2 different triples: u1 = (x1,y1,z1)T, u2 = (x2,y2,z2)T, such that:

Au1 = Au2 = w = (0,-1)T.

one way to do this, is to solve the linear system of equations:

x + 2y = 0
y - z = -1

if you do this correctly, you should wind up with a "free variable". two different choices for this free variable will give you two different u1, u2.

(as for the "R" notation, R2 means simply the set of pairs of real numbers, and R3 is the set of trios (triples) of real numbers. these are given a notion of vector addition by "adding coordinate-by-coordinate" and a notion of scalar multiplication by "scaling each coordinate by a real number" that is, multiplying each coordinate by the same real number).
 

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