Question about Matrix Linear Transformation

albaboi123
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i'm studying for my midterm and I'm stumped on this section about Lienar Transformations...hope u guys can help

Homework Statement



question goes something like this
1) Find the standard matrix for the linear operator define by the equations (which is easy)
and then determine wheter the operator is one-to-one and or onto

example question is
w1=-x+3y+2z
w2=2x+4z
w3=x+3y+6z

2) Show that the range of the linear operator defined by the equations
w1=4x-2y
w2=2x-y
is not all of R^2, and find a vector that's not in the range.




The Attempt at a Solution



1) so the standard matrix is

[-1 3 2]
[2 0 4]
[1 3 6]

now how do i tell if its 1-1 and/or onto?
 
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Put the matrix into reduced row echelon form. If every row has a pivot, then the vectors are a basis for the codomain and the transformation is onto. The transformation is one-to-one if all the vectors are linearly independent. Since this is a 3x3 matrix, either the function will be both one-to-one and onto, or the transformation will be neither.
 
A linear transformation is one-to-one and onto if and only if it has an inverse linear transformation. When it is written as a matrix, this is true if and only if the determinant of the matrix is not 0.
 
Also, since it is an operator on R^3 it's injective and onto iff the null space is trivial

for a linear operator A the following are equivalent: A is invertible, A is injective, A is onto

since it is an operator you need only consider if the null space is trivial, since if it is trivial it is injective and hence bijective, if it is not then it is neither injective nor onto
 
Last edited:
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...

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