Let $P_m$ be space of polynomials of degree $\leq m$ with real coefficients. Is it surjective? Let $P_m$ be space of polynomials of degree $\leq m$ with real coefficients. Consider the application
$$L:P_3 \to P_3, L(g)=(xg)^{'}$$
(i.e., L is the derivative of the product xg). Show that L is linear. Is it injective and surjective? Find a matrix representation for L with respect to the standard basis $\{ 1, x, x^2, x^3\}$.
Show that L is linear: A map $L:P_3 \to P_3$ is linear if for all $g,h \in P_3$, $\lambda \in \textbf{R}$ holds:


*

*$L(g+h)=L(g)+L(h)$

*$L(\lambda g)=\lambda L(g)$


Let $g \in P_3$ and $h \in P_3$.
\begin{equation*}
\begin{split}
L(g+h) & = (x(g+h))' \\
& =(xg+xh)' \\
& =(xg)' + (xh)' \\
& =L(g) + L(h) \\
\end{split}
\end{equation*}
Let $g \in P_3$ and $\lambda \in \textbf{R}$.
\begin{equation*}
\begin{split}
L(\lambda g) & =(\lambda xg)' \\
& =\lambda (xg)' \\
& =\lambda L(g)
\end{split}
\end{equation*}
Hence L is linear.
Is it injective and surjective?
Important Definitions:


*

*A linear operation T is called injective if $x=0$ whenever $Tx=0$. (i.e. If $Tx=0$ then $x=0$.)

*It is called surjective if the range of T, defined $range(T)=\{Tx:x\in \nu\}=T\nu$, is equal to the entire vector space W.


Claim: L isn't injective.
Proof (by contradition): Assume L is injective Using the definition, above we must show that if $L(g)=0$, then $g=0$. Let $ g \in P_3$ by the linear application L we have:
$$L(g)=(xg)'=0$$
Since $g \in P_3$, g could be equal to 0 and then we are done. But what if x=0, well then g can be anything, of course excluding $g=0$, and we will get $L(g)=0$ and $g \neq 0$. Hence L isn't injective all the time.
Is L surjective? Why?
Find a matrix representation for L with respect to the standard basis $\{1,x,x^2,x^3\}$
I know how to find it when it comes to normal polynomials. How do I do it for the derivative?
 A: Is it injective?
To show that a linear function $\phi$ is injective, it is sufficient to show that $\phi(x) = 0 \iff x = 0$ - because if $\phi(x) = \phi(y)$, then $\phi(x-y) =0$, so $\phi$ is injective means that we must have $x=y$.
If $L(g) = (xg)' = 0$, then by integrating, $$xg = C$$ for some constant $C$. Can $g$ be non-zero?
Is it surjective?
Let $f$ be a polynomial of degree $\le 3$. We want to find $g$ such that $L(g)=f$. But then $$xg = \int f$$
Can we use this to find an appropriate $g$? Be careful - it's clear that we can find a function $g$, but you need to show that it is a polynomial of degree $\le 3$.
Note that in this context $x$ is a formal symbol representing the polynomial $P(x) =x$. It can't take any particular value - it doesn't make sense to say $x=0$ here.
Find a matrix representation for $L$ with respect to the standard basis $\{1,x,x^2,x^3\}$:
The $i^{\mathrm{th}}$ column of a matrix is the image of the $i^{\mathrm{th}}$ basis vector in terms of the basis. Can you find the image of $1,x ,x^2$ and $x^3$ under $L$? Can you express these in terms of the basis $\{1,x,x^2,x^3\}$?
For example, $L(x) = (x\cdot x)' = (x^2)' = 2x$, which is $(0,2,0,0)$ in terms of the basis $\{1,x,x^2,x^3\}$. So the second column of your matrix will be:
$$\left(\begin{matrix}
* & 0 & * & *\\
* & 2 & * & *\\
* & 0 & * & *\\
* & 0 & * & *\\
\end{matrix}\right)$$
