Linear Algebra: Matrix Transformations

skylit
Messages
6
Reaction score
0

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 = <br /> \begin{pmatrix}<br /> 1 &amp; 2 &amp; 0\\<br /> 0 &amp; 1 &amp; -1\\<br /> \end{pmatrix}<br />
w=
<br /> \begin{pmatrix}<br /> 0\\<br /> -1\\<br /> \end{pmatrix}<br />

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 = <br /> \begin{pmatrix}<br /> x\\<br /> y\\<br /> z\\<br /> \end{pmatrix}<br /> ?

Detailed instruction would be much appreciated, as I am anxious to grasp the subject matter.
 
Physics news on Phys.org
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).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top