Finding Bases for Kernel and Image of Linear Transformation I'm trying to solve a linear transformation problem.
Let $ \psi: \mathbb R_3 [x] \to \mathbb R_4 [x] $ be defined by $ \psi : p(x) \mapsto x^4 p(1/x)+p(x)$
Q) Show that $\psi$ is a linear transformation and find the bases for ker($\psi$) and Im($\psi$).
I know that to show it's a linear transformation you must show it satisfies the conditions:
$\phi(x+y)=\phi(x)+\phi(y) $ and $ \phi(\lambda x)= \lambda \phi(x)$ but don't know how to go about solving it.
Your help will be greatly appreciated...     
 A: Let's think about what $\psi$ does to a few polynomials.
Say $p(x) = x^3 -2x$. Then $p(1/x) = \frac{1}{x^3} - 2\frac{1}{x}$, so 
$$x^4p(1/x) = x^4\left(\frac{1}{x^3}-2\frac{1}{x}\right) = x - 2x^3.$$
In particular, note that it is a polynomial, even though the computation involved something which is not a polynomial. Since $p(x)$ has degree at most $3$, then the highest power of $x$ that can occur in the denominator of $p(1/x)$ is $x^3$, so multiplying by $x^4$ will guarantee all powers of $x$ that occur are nonnegative. You should also try to make sure that the image always has degree at most $4$ (so that it actually is in $\mathbb{R}_4[x]$).
So, if $p(x) = x^3-2x$, for example, then
$$\psi(x) = x^4p(1/x) + p(x) = (x-2x^3) + (x^3-2x) = -x^3 -x.$$
If $p(x)$ and $q(x)$ are polynomials, then remember that $(p+q)(u) = p(u)+q(u)$. So $(p+q)(1/x) = p(1/x) + q(1/x)$, whatever $p(1/x)$ and $q(1/x)$ might be. So, what is $\psi(p+q)$? It is
$$\psi(p+q) = x^4\Bigl( (p+q)(1/x)\Bigr) + (p+q)(x) = x^4\Bigl(p(1/x)+q(1/x)\Bigr) + p(x)+q(x).$$
Is it equal to $\psi(p) + \psi(q)$? Well, evaluate and check.
If $p(x)$ is a polynomial and $\alpha$ is a scalar, then $\alpha p$ is the polynomial which, when evaluated at $u$, gives $\alpha p(u)$. So
$$\psi(\alpha p) = x^4\Bigl(( \alpha p)(1/x)\Bigr) + (\alpha p)(x) = x^4\alpha p(1/x) + \alpha p(x).$$
What is $\alpha\psi(p)$? Is it equal to $\psi(\alpha p)$?
Once you know that it is a linear transformation (or before, if you want to try to get a better handle on it) see if you can figure out a formula for $\psi(a_0 + a_1x + a_2x^2+a_3x^3)$ in terms of $a_0$, $a_1$, $a_2$, and $a_3$ (equivalently, if you know it is linear, figure out $\psi(1)$, $\psi(x)$, $\psi(x^2)$, and $\psi(x^3)$; why will that be enough?) This should help you find out what the image is, and so a basis for it; and what the kernel is (and so a basis for it).
