Exercise C43

Define the following: \(T:p_{3} \rightarrow p_{2}\) by \(T(a+bx+cx^{2}+dx^{3})=b+2cx+3dx^{2}\). Find thr Pre-Image of 0. Does this linear transformation seem familiar?

Solution

What the pre-image is ?

Let \(f:A\rightarrow B\) be a map between sets A and B. Let \(Y \subseteq B\). Then the preimage of Y under f is denoted by \(f^{-1}(Y)\), and is the set of all elements of A that map to elements in Y under f. Thus

\(f^{-1}(Y)={a \in A|f(a) \in Y}\)

One is not to be mislead by the notation into thinking of the preimage as having to do with an inverse of f. The preimage is defined whether f has an inverse or not. Note however that if f does have an inverse, then the preimage \(f^{-1}(Y)\) is exactly the image of Y under the inverse map, thus justifying the perhaps slightly misleading notation.

For any Y subset= B, it is true that

\(f(f^{-1}(Y)) \subseteq Y\)

with equality occurring, if f is surjective, and for any subset X subset= A, it is true that

\(X \subseteq f^{-1}(f(X))\) with equality occurring if f is injective.

Preimages occur in a variety of subjects, the most persistent of these being topology, where a map is continuous, by definition, if the preimage of every open set is open.

Preimage of T \[ T^{-1}(0) \]

The preimage of T is the set of all polynomials where the following holds true \[ T(a+bx+cx^{2}+dx^{3})=0 \]

This implies that the following is also true \[ b+2cx+3dx^{2}=0 \] Therefore it can be said that zero represents the 0 polynomial, in otherwords the set of all polynomials where a=0,b=0,and c=0 thus polynomials of degree 0.

How does this transformation look familiar? \[ T(a+bx+cx^{2}+dx^{3})=b+2cx+3dx^{2} \] Is the same as \[ \frac{dx}{dy}(a+bx+cx^{2}+dx^{3})=b+2cx+3dx^{2} \]

In other words, the expression on the right is the derivative of the polynomial on the left assuming a, b, and c are non zero constants. The derivative of a constant is zero. The derivative of the other terms can be found using the power rule. \[ \frac{dx}{dy}(ax^{n})=n.ax^{n-1} \]