# Thread: Showing that a function is a linear map

1. ## Showing that a function is a linear map

Show that the function $\displaystyle T : \mathbb{P}_3(\mathbb{R})\rightarrow \mathbb{P}_3(\mathbb{R})$ defined by

$\displaystyle T(p)=4p'+3p$ where $\displaystyle p'(x)=\frac{dp}{dx}$

is a linear map

Would I begin the question like the following:

$\displaystyle p(x)=a_0+a_1x+a_2x^2+a_3x^3\Rightarrow p'(x)=a_1+2a_2x+3a_3x^2$

$\displaystyle T\begin{pmatrix}a_0\\a_1\\a_2\\a_3\end{pmatrix}=4\ begin{pmatrix}a_1\\2a_2\\3a_3\\0\end{pmatrix}+3\be gin{pmatrix}a_0\\a_1\\a_2\\a_3\end{pmatrix}$

2. That seems more tedious than necessary. Just use the definition of "linear map".

Suppose p and q are two polynomials in $\displaystyle \mathbb{P}_3(\mathbb{R})$ and a is a real number.

What is T(p+ q)? Can you show that is the same as T(p)+ T(q)? What is T(ap)? Can you show that is the same as aT(p)?
Use basic properties of the derivative.

3. Originally Posted by HallsofIvy
That seems more tedious than necessary. Just use the definition of "linear map".

Suppose p and q are two polynomials in $\displaystyle \mathbb{P}_3(\mathbb{R})$ and a is a real number.

What is T(p+ q)? Can you show that is the same as T(p)+ T(q)? What is T(ap)? Can you show that is the same as aT(p)?
Use basic properties of the derivative.
By the way, was what I was doing before still correct?

Is this proof good enough

$\displaystyle T(p+q)=4(p+q)'+3(p+q)$

Now using properties of derivatives and functions

$\displaystyle =4p'+q'+3p+3q$

$\displaystyle =(4p'+3p)+(q'+3q)$

$\displaystyle =T(p)+T(q)\Rightarrow \textrm{Addition condition satisfied}$

$\displaystyle T(ap)=4(ap)'+3(ap)$

$\displaystyle =4ap'+3ap$

$\displaystyle =a(4p'+3p)$

$\displaystyle =a(T(a))\Rightarrow \textrm{Scalar multiplication condition}$

Therefore T is a linear map