# Please explain, this is a Linear transformation or not ?

## The second derivative of a four degree polynomial is a Linear Transformation or not?

Jun 26, 2018

See below

#### Explanation:

A trasformation $T : V \setminus \to W$ is said to be linear if it has the following two properties:

• $T \left({v}_{1} + {v}_{2}\right) = T \left({v}_{1}\right) + T \left({v}_{2}\right)$ for every ${v}_{1} , {v}_{2} \setminus \in V$
• $T \left(c v\right) = c T \left(v\right)$ for every $v \in V$ and every scalar $c$

Note that the second property assumes that $V$ is embedded with two operations of sum and scalar multiplication. In our case, the sum is the sum between polynomials, and the multiplication is the multiplication with real numbers (I assume).

When you derive a polynomial you lower its degree by $1$, so if your derive a polynomial of degree $4$ twice, you will get a polynomial of degree $2$. Note that, when we speak of the set of all four degree polyinomial, we actually mean the set of all polynomials of degree at most four. In fact, a generic degree four polynomial is

${a}_{0} + {a}_{1} x + {a}_{2} {x}^{2} + {a}_{3} {x}^{3} + {a}_{4} {x}^{4}$

If you want the degree two polynomial $3 + 6 x - 5 {x}^{2}$, for example, you simply choose

${a}_{0} = 3 , {a}_{1} = 6 , {a}_{2} = - 5 , {a}_{3} = {a}_{4} = 0$

With that being said, let's identify the polynomial space of degree $n$ with ${P}_{n}$, and define our operator $T : {P}_{4} \setminus \to {P}_{2}$ such that $T \left(f \left(x\right)\right) = f ' ' \left(x\right)$

Let's proove the first property: assume we have the polynomials

${p}_{1} = {a}_{0} + {a}_{1} x + {a}_{2} {x}^{2} + {a}_{3} {x}^{3} + {a}_{4} {x}^{4}$

and

${p}_{2} = {b}_{0} + {b}_{1} x + {b}_{2} {x}^{2} + {b}_{3} {x}^{3} + {b}_{4} {x}^{4}$

This means that ${p}_{1} + {p}_{2}$ equals

$\left({a}_{0} + {b}_{0}\right) + \left({a}_{1} + {b}_{1}\right) x + \left({a}_{2} + {b}_{2}\right) {x}^{2} + \left({a}_{3} + {b}_{3}\right) {x}^{3} + \left({a}_{4} + {b}_{4}\right) {x}^{4}$

$T \left({p}_{1} + {p}_{2}\right)$ is the second derivative of this polynomial, so it is

$2 \left({a}_{2} + {b}_{2}\right) + 6 \left({a}_{3} + {b}_{3}\right) x + 12 \left({a}_{4} + {b}_{4}\right) {x}^{2}$

(I applied twice the power rule for derivation: the second derivative of ${x}^{n}$ is $n \left(n - 1\right) {x}^{n - 2}$)

Now let's compute $T \left({p}_{1}\right)$, i.e. the second derivative of ${p}_{1}$:

$2 {a}_{2} + 6 {a}_{3} x + 12 {a}_{4} {x}^{2}$

Similarly, $T \left({p}_{2}\right)$, i.e. the second derivative of ${p}_{2}$, is

$2 {b}_{2} + 6 {b}_{3} x + 12 {b}_{4} {x}^{2}$

If you sum these expression, you can see that we have

$T \left({p}_{1} + {p}_{2}\right) = T \left({p}_{1}\right) + T \left({p}_{2}\right)$

The second property is shown in a similar fashion: given a polynomial

$p = {a}_{0} + {a}_{1} x + {a}_{2} {x}^{2} + {a}_{3} {x}^{3} + {a}_{4} {x}^{4}$

we have, for any real number $c$,

$c p = c {a}_{0} + c {a}_{1} x + c {a}_{2} {x}^{2} + c {a}_{3} {x}^{3} + c {a}_{4} {x}^{4}$

its second derivative is thus

$2 c {a}_{2} + 6 c {a}_{3} x + 12 c {a}_{4} {x}^{2}$

which again is the same as computing $T \left(p\right)$, and then multiply everything by $c$, i.e. $T \left(c p\right) = c T \left(p\right)$