How to prove $(Tp)(x) = x^2p(x)$ is a linear map?

In Linear Algebra Done Right, it gives an example of linear maps as follow:

multiplication by $$x^2$$

Define $$T\in L(P(R),P(R))$$ by $$(Tp)(x) = x^2p(x)$$ for $$x \in \mathbb{R}$$

My attempt: $$(Tp)(x) = x^2p(x)$$ $$(Tp)(x+y) = x^2p(x+y) = x^2p(x) + x^2p(y) = (Tp)(x) + (Tp)(y)$$ I am not sure whether I can do $$p(x+y) = p(x) + p(y)$$. It seems I cannot. Right?

$$(Tp)(\lambda x) = x^2p(\lambda x) = \lambda x^2p(x) = \lambda(Tp)(x)$$ I am also not sure whether I can do $$p(\lambda x) = \lambda p(x)$$.

• What's $P(R)$? Polynomials with coefficients in $\Bbb R$? – Zachary Selk Jan 12 at 1:03
• @ZacharySelk I think it is. – JOHN Jan 12 at 1:04
• If so your attempt is very wrong. – Zachary Selk Jan 12 at 1:04
• You're on a wrong track. You have to think about $(T(p+q))(x)$, not $(Tp)(x+y)$. The linearity is about the addition and scalar multiplication in the space of polynomials, not in the real numbers where you are evaluating the polynomials. – Ethan Bolker Jan 12 at 1:05

Remember, you are adding polynomials, not independent variables like $$x$$ and $$y$$. If you want to check additivity, you need to verify that, given polynomials $$p$$ and $$q$$, that $$T(p + q) = Tp + Tq.$$ Now, $$T(p + q), T(p)$$, and $$T(q)$$ are all polynomials themselves, so if you want to show that they are equal, you can show that they are equal at some arbitrary point $$x$$. That is, show $$(T(p + q))(x) = (Tp)(x) + (Tq)(x).$$ Note that, by definition of $$T$$, we have that $$(Tp)(x) = x^2 p(x)$$ for all $$x$$, and similar for $$q$$. So, equivalently, $$x^2(p + q)(x) = x^2 p(x) + x^2 q(x).$$ All I've done is use the definition of $$T$$. Then, we need to use the definition of $$+$$ for polynomials, which is $$(p + q)(x) = p(x) + q(x)$$. Note the $$+$$ on the left is the $$+$$ being defined, and adds functions. The $$+$$ on the right is the usual addition between real scalars, as $$p(x)$$ and $$q(x)$$ are actually just real numbers (although, unknown real numbers, since we don't know $$x$$, or $$p$$ and $$q$$). Therefore, we have $$x^2 (p + q)(x) = x^2(p(x) + q(x)) = x^2p(x) + x^2q(x),$$ using the definition of polynomial addition, and the distributive law of real numbers.

Have another go at scalar multiplication. Remember the definition of polynomial scalar multiplication: $$(\lambda \cdot p)(x) = \lambda \cdot p(x)$$ Again, on the left is the operation being defined, between a real number and a real valued polynomial. The operation on the right is ordinary multiplication between real numbers.

You are mistaken on the definition of $$T$$. Since $$T$$ is an application from $$P(R)$$ into $$P(R)$$ (I assume that $$P(R)$$ stands for the vector space of polynomials with real coefficients), the input is a polynomial $$p$$ and the output is also a polynomial function $$T(p)$$, usually written as $$Tp$$.

What you have to show in order to prove that $$T$$ is linear is: 1. for any $$p$$ and $$q$$ in $$P(R)$$, $$T(p+q)=T(p)+T(q)$$. 2. for any $$p$$in $$P(R)$$ and any real number $$c$$, $$T(cp)=cT(p)$$.

Then... what is that $$x$$ in the definition of $$T$$? Since $$f=T(p)$$ is itself a function, it is defined by expliciting its evaluation $$f(x)$$ at every real number $$x$$.

To sum up, $$T$$ is an application (= a function) that transforms a polynomial function $$p$$ into a new ploynomial application $$f=T(p)$$.

Let's prove the point $$1$$ above (point $$2$$ is proved similarly). Let $$p$$ and $$q$$ be two polynomials. Then, we need to check the equality of functions $$T(p+q)=T(p)+T(q)$$. For, consider a real number $$x$$. We have

$$T(p+q)(x)=x^2(p+q)(x)=x^2(p(x)+q(x))=x^2p(x)+x^2q(x) = T(p)(x) + T(q)(x)$$

that is $$T(p+q)(x) = (T(p)+T(q))(x)$$. The claim $$T(p+q)=T(p)+T(q)$$ is proved! Make sure you understand each step of the computation above, this kind of game with definitions/formulas is important for abstract algebra.

Note: the notation $$T(p)(x)$$ means that the function $$T(p)$$ is evaluated at $$x$$. If you let $$f=T(p)$$, then it is just $$f(x)$$.

Note 2: in general, polynomials are not linear applications, so we don't have $$p(x+y)=p(x)+p(y)$$ and $$p(cx)=cp(x)$$.

Note 3: for simplicity, I identified polynomials and polynomial functions, though they are not exactly the same. It is fine because we are considering real coefficients here.