# The Range and Nullspace of the Linear Transformation $T (f) (x) = x f(x)$ ## Problem 672

For an integer $n > 0$, let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the standard basis.

Let $T : \mathrm{P}_n \rightarrow \mathrm{P}_{n+1}$ be the map defined by, for $f \in \mathrm{P}_n$,
$T (f) (x) = x f(x).$

Prove that $T$ is a linear transformation, and find its range and nullspace. Add to solve later

## Solution.

### $T$ is a linear transformation

We must show that for polynomials $f, g \in \mathrm{P}_n$ and scalar $c \in \mathbb{R}$, the function $T$ satisfies $T(f+g) = T(f) + T(g)$ and $T(cf) = c T(f)$. The first is checked using associativity of multiplication:
$T(f+g)(x) = x (f+g)(x) = x f(x) + x g(x) = T(f)(x) + T(g)(x) .$ Similarly,
$T(cf)(x) = x (cf)(x) = c x f(x) = c T(f)(x).$

### The nullspace of $T$

The nullspace of $T$ is the set of polynomials $f(x)$ such that $T(f) = 0$. That is, $x f(x) = 0$. But this product can only be $0$ if one of the terms being multiplied is $0$. Because $x \neq 0$, we must have that $f(x) = 0$. Thus the nullspace is the trivial vector subspace $\{ 0 \}$.

### The range of $T$

The range of $T$ is the set of polynomials of the form $x f(x)$ for an arbitrary polynomial $f \in \mathrm{P}_n$. We can calculate this by calculating $T$ for a basis of $\mathrm{P}_n$. Let $B = \{ 1 , x , x^2 , \cdots , x^n \}$ be a basis of $\mathrm{P}_n$.

We see that $T(x^i) = x^{i+1}$ for $0 \leq i \leq n$, and so the image of the basis is the set
$T( B ) = \{ x , x^2 , x^3 , \cdots , x^{n+1} \}.$ Then the range of $T$ must be the span of this set. Specifically,
$\mathcal{R} ( T ) = \Span ( T( B ) ) = \Span ( x , x^2 , x^3 , \cdots , x^{n+1} ).$ Add to solve later

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