# Find the image and kernel of the linear transformation.

Find the image and kernel of the linear transformation.

$$T: P_{1}\rightarrow P_{2} , T(p(x))=xp(x)+p(0)$$?

I'm trying to solve this exercise, but I'm stuck with the linear transformation notation, I think is just a polynomial transformation that increments the polynomial degree by one, but then I don't know how to represent it for getting the image and kernel.

• To be clear, do you need help figuring out what these spaces are, or do you just need help in writing them in an acceptable form? If the latter, I suggest writing them as a span of a basis (or calling it "trivial" if it contains only the $0$ vector). May 30, 2019 at 5:20

You see, $$T(.)$$ maps one polynomial space - which is by the way a vector space - in another. Oh, and not only does $$T(.)$$ increases the degree of the polynomial by one, it also sums it with its vertical intercept.
To find the kernel of $$T(.)$$ you'll have to figure out what group of polynomials that get mapped to the origin by it.
1. Note that $$ker(T)=\{p(x)\in P_1:T(p(x))=0\}=\{p(x)\in P_1:xp(x)+p(0)=0\}$$ So the kernel is exact the set of polynomials $$p(x)$$ that satisfy the equation $$xp(x)+p(0)=0$$ You may not have encountered equations with polynomials being solutions. Nonetheless, keep calm and do the question: can you see that the only polynomial that satisfies the equation is $$0$$ and hence $$ker(T)=\{0\}$$?
2. For the image, or range, note that $$range(T)=\{T(p(x))\in P_2|p(x)\in P_1\}=\{xp(x)+p(0)|p(x)\in P_1 \}$$. Hence the image of $$T$$ is exactly the polynomials in the form $$xp(x)+p(0)$$ where p(x) is some other polynomials. Depending on the requirement of your question, you may have to further identify the exact set of polynomials that have the above form. Can you see that the polynomials in the above form are exactly those polynomials having the same coefficients for $$x^1$$ and $$x^0$$?