Linear algebra transformations, kernel, range and confusion

Question

I have a vector space $V$ of polynomials in the variable $x \in \mathbb R$. The transformation $f$ is defined as follows:

$ f : V → V : p(x) → x^2 \left(\frac{d^2 p(x)}{dx^2}\right)$

i.e.: deriviate twice in x and multiply by $x^2$.

And now the question:

Give the exact description of all elements of the kernel of f in the range of f. What are the dimensions of V, ther kernel of f and the range of f?

This is where I'm completely stuck; how is this even a linear transformation if it involves squares and how do I find this kernel and the matrix A that corresponds with f? I need this matrix because I need to find the eigenvalues and eigenvectors although the assignment states that this is a theoretical exercise and that matrices are unnecessary here.

Answer 1

Let me denote the operator by $T$, and the second derivative by $D^2$.

Note that the derivative is a linear operator, so for $p,q \in V$ and $a, b \in \mathbb R$, you have $$ T(a p(x)+ b q(x)) = x^2 D^2 (ap(x)+bq(x)) = a x^2 D^2 p(x) + b x^2 D^2 q(x) = a Tp(x) + bTq(x),$$ and $T$ is a linear operator.

To find the kernel, you should determine for which $p\in V$ you have $$ Tp(x) = x^2 D^2 p(x) = 0 \quad \iff \quad D^2p(x) = 0. $$ Note that this can be solved by simply integrating twice.

To find the range, you should check for which $q \in V$, there is a $p \in V$ such that $$ Tp(x) = q(x) \quad \iff \quad x^2 D^2p(x) = q(x). $$ Note that $q$ is a polynomial and that the solution $p$ to this equation should be a polynomial.

Answer 2

$V$ is the vector space of polynomials. Think about $V$ as infinite real sequences such that all but finitely many elements are $0$. (The identification being: the list of coefficients $a_0, a_1, \ldots$ of the polynomial.) Then multiplication by $x^2$ is just a transformation that maps a sequence $s$ to another one that is obtained from $s$ by writing two zeroes in the beginning of the sequence. (More precisely, shift the sequence by 2 to the right and write $0$ into the term $a_0$ and $a_1$.) This is clearly a linear operation. The same holds for derivation: it just multiplies every $a_i$ by $i$, and then deletes the first term (whose index is zero.)

Answer 3

Note that both $p\mapsto dp/dx$ and $p\mapsto x\cdot p$ are $\Bbb R$-linear maps, and $f$ is a composition of these maps.

For the kernel, we have $f(p)=0$ iff $d^2p/dx^2=0$, i.e. iff $p$ is at most linear.

For the range, we have $q=f(p)$ iff $q$ is divisible by $x^2$, i.e. has zero constant term and zero coefficient for $x$.

For the eigenvectors, write $p=a_0+a_1x+\dots +a_nx^n$.

Answer 4

One way to view vectors in this space is to convert each polynomial into a column vector filled with its coefficients, so the first component is the constant of the polynomial, the second component is the coefficient for the $x$ term, the third component is the coefficient for the $x^2$ term, and so on. So multiplying by $x^2$ basically just shifts each coefficient down two components and fills the first two components with 0's, which is a linear transformation $T(v)$ because $T(av+bw)=a(T(v)+b(T(w))$ for constants $a,b$ and vectors $v,w$.

The kernel is just all polynomials which map to the $0$ vector after the transformation. The "multiplying by $x^2$" part will never produce a $0$ vector unless $p(x)$ is the $0$ vector, so the second derivative of $p(x)$ has to be 0. Therefore the kernel of the given transformation is the set of all polynomials of degree 1, so polynomials of the form $a+bx$ for constants $a,b$.

Now, the range of the transformation is every polynomial which can be obtained through the transformation. The zero vector is included, since the kernel is non-empty. The other polynomials possible are any polynomial of degree at least 2 because we multiply by $x^2$ after the derivative.