Patrick G. answered 05/31/21
Physics Degree with 4+ years of tutoring experience
First, we will tackle the image of v under this transformation, T. The image is going to describe the set of all possible outputs for a given set of inputs under a given transformations. In layman's terms, we take all of the given inputs, and plug it into the function. Whatever we get out is our image.
For the transformation above, T(v1,v2,v3) = (4v2 − v1, 4v1 + 5v2). To find the image of v, we plug v into T.
T(v)=T(3,-4,-1) = (4*(-4)-(3),4(3)+5(-4))
We then simplify to get our final answer
Im(v)=(-19,-8)
A preimage is the set of all inputs that result in a given input under a given transformations. In layman's terms, we will take our output (w) and see what all of the possible inputs are that result in our output.
Let P be our preimage...
P=(p1,p2,p3)
T(P)=w=(6,18)
(6,18)=(4p2-p1,4p1+5p2)
Breaking this up into a system of two equations...
6=4p2-p1
18=4p1+5p2
We can now solve the system of equations for p1 and p2 to get the following...
p1=2, p2=2
p3 does not appear anywhere in the system of equations, thus it is a free variable. This implies any vector of the form (2,2,p3) will map to w under T. (T(2,2,p3)=w for all possible p3). Thus we can say the following is our preimage...
preimage(w)={(2,2,p3) | p3 is an element of all real numbers}