Linear Algebra: Matrix Transformations

[skylit]

Homework Statement

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

A = $$\begin{pmatrix} 1 & 2 & 0\\ 0 & 1 & -1\\ \end{pmatrix}$$
w=
$$\begin{pmatrix} 0\\ -1\\ \end{pmatrix}$$

Homework Equations

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

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 $R$, either.

But from inferring from the problem, is it safe to assume that $w = Au$?
Also, would I have to use something along the lines of
u = $$\begin{pmatrix} x\\ y\\ z\\ \end{pmatrix}$$ ?

Detailed instruction would be much appreciated, as I am anxious to grasp the subject matter.

 

[Deveno]

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).