Let g: $P_2$ → $P_3$ be defined as follows: $g(a_2x^2 + a_1x + a_0) = (a_2 − a_1)x_3 − a_1x + 9a_0$
Find the kernel and range of g. Give bases for these subspaces as comma-separated lists (e.g. a basis for $ℝ^3$ is {(1,0,0,), (0,1,0), (0,0,1)} ).
I started with the kernel setting the output equal to $0$
$g(a_2x^2 + a_1x + a_0) = (a_2 − a_1)x^3 − a_1x + 9a_0 = 0$
And then I can set the coefficients on x equal to $0$
$(a_2 − a_1)=0; −a_1=0; a_0 = 0$
And I get
$a_2 = a_1; a_1=0; a_0 = 0$
Looking back at the original output of the transformation it looks like that means all $a_i$ must be $0$ in order to get $(a_2 − a_1)x^3 − a_1x + 9a_0 = 0$ for all values of x
But since $a_2 = a_1$, I'm not sure if that changes anything, or what my dimension is.
But I don't know how to express it as a basis.
Since g is $P_2$ → $P_3$, I'm expecting a 2D vector to be part of my kernel basis but {(0,0)} didn't seem to work
From example 2.2 on this page they say $x-1$ is a basis. Obviously it's a different question but I don't know how to proceed.
Wtht with the range, I think I can do a similar thing if someone shows me how to get the kernel
Best Answer
You are on the right track. With $a_2 = a_1 = a_0 = 0$, you have shown that the kernel of $g$ is the set of all polynomials of degree 2 or less with coefficients equal to zero. That is,
$$\ker(g) = \{0\}.$$
There is a very special basis for this space - $\{0\}$ - and we define the dimension of this space to be zero.
Now, using the rank-nullity theorem (or dimension theorem or whatever else your book or notes may call it), it follows that the dimension of the range of $g$ (or rank of $g$) is $3$. Why? This follows from the dimension of the domain being 3 and the dimension of the kernel of $g$ being $0$.
Since it is clear that no quadratic coefficient is present in the range, it follows that the range of $g$ is
$$\{ b_3x^3 + b_1x + b_0 | b_i \in \mathbb{F} \}$$
where $\mathbb{F}$ is whatever field that you are given here (the real numbers, the complex numbers, the rational numbers, etc.).
At this point, it is easy to see that $\{1, x, x^3\}$ is a basis for the range of $g$.