I'm working on the following problem, not entirely sure if what I'm doing is correct.
Let $F$ be a field, and let $V={f(x)∈F[x] : deg(f)≤2}$ be the vector space of quadratic polynomials. Consider the linear operator $A:f(x)→f(x+ 2)−f(x)$ from $V$ to itself.
a) Write the matrix of A with respect to the standard basis {$1, x, x^2$} of $V$.
b) Describe the kernel of $A$.
Here is what I have:
a) Seeing as A is a linear operator, we can say that $f(x+2) - f(x) = f(2)$, so $A$ simply evaluates a given function at $2$.
Thus, $A(1) = 1$, $A(x) = 2$, and $A(x^2) = 4$. Expressing these results as combinations of the basis gives us the 3x3 matrix with a top row of 1, 2, 4 and zeroes everywhere else.
b) The kernel of A will consist of all quadratic polynomials who equal to zero when evaluated at 2. (Not sure if I need to be more specific here?)
Thanks
Not sure that this is correct $f(x+2)-f(x)=f(2)$ since $f$ is not linear. Take $f(x)=x^2$ then $f(x+2)-f(x)=4x+4$
But taking in for $f_0(x)=1$ $f_1(x)=x$ and $f_2(x)=x^2$ we get that $A(f_0)=0$ $A(f_1)=2$ and $A(f_2)=4f_1+4f_0$. So the matrix is:
$$\left(\begin{array}{ccc} 0 & 0 & 0\\ 2 & 0 & 0\\ 4 & 4 & 0\end{array}\right)$$